Click to Chat
1800-2000-838
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Motion Under Gravity Whenever a body is released frrom a height, it travels vertically downward towards the surface of earth.This is due to the force of gravitational attraction exerted on body by the eareth. The acceleration produced by this force is called acceleration due to gravity and is denoted by ‘g’. Value of ‘g’ on the surface of earth is taken to the 9.8 m/s^{2} and it is same for all the bodies. It means all bodies (whether an iron ball or a piece of paper), when dropped (u=0) from same height shoulfd fall with same rapidity and should take same time to reach the earth. Our daily observation is contrary to this concept. We find that iron ball falls more rapidlly than piece of paper. This is due to the presence of air which offers different resistance to them. In the absence of air both would have taken same time to reach the surface of earth. When a body is dropped from some height (earth's radius = 6400 km), it falls freely under gravity with constant acceleration g (= 9.8 m/s^{2}) provided the air resistance is negligible small. The same set of three equations of kinematics (where the acceleration remains constant) are used in solving such motion. Here, we replace by and choose the direction of y-axis conveniently. When the y-axis is chosen positive along vertically downward direction, we take as positive and use the equation as v = u + gt, v^{2} = u^{2} + 2gh, and h = ut + 1/2gt^{2} where h is the displacement of the body and u is initial velocity of projection in the vertically downward direction. However, if an object is projected vertically upward with initial velocity u, we can take y-axis positive in the vertically upward direction and the set of equations reduces to v = u - gt, v^{2} = u^{2} - 2gh, and h = ut - 1/2gt^{2} In order to avoid confusion in selecting as positive or negative, it is advisable to take the y-axis as positive along vertically upward direction and point of projection as the origin. We can now write the set of three equations in the vector form: , and where h is the displacement of the body. Problem: The motion of a particle is described by the equation u = at. The distance travelled by the particle in the first 4 second. Solution: Because for the motion u = at. So acceleration is uniform which is equal to a. Therefore, Distance traveled = 1/2[(a)(4)^{2}] = 8a Problem: A body moving with a constant retardation in straight line travels 5.7 m and 3.9 m in the 6th and 9th second, respectively. When will the body come momentarily to rest? Solution: A body moving with initial velocity u and acceleration a, traverses distance S_{n} in n^{th}second of its motion. S_{n} = u + (1/2)(2n - 1)a => 5.7 = u + (1/2)(2 x 6 - 1) a or 5.7 = u + (11/2) a and 3.9 = u + (1/2)(2 x 9 - 1)a or, 3.9 = u + (17/2) a Solving eqns. (1) and (2) we get, u = 9 m/s and a = -0.6 m/s^{2}. If the body stops moving after t seconds, then from the relation v=u+at Thus, 0 = 9 + (-0.6)t or, t = (9/0.6)s = 15s Problem (JEE Advanced): A stone, thrown up is caught by the thrower after 6s. How high did it go and where was it 4 s after start? g = 9.8 m/s^{2} Solution: Time to go up and come back = 6s Thus, time to reach the highest point = (6/2) s = 3s From point of projection to the highest point we have u = ?, v = 0, a = -9.8 m/s^{2}, t = 3s Using the relation, v = u + at 0= u – 9.83 Thus, u = 29.4 m/s^{2} Maximum height, H = u^{2}/2g = [(29.4)^{2}/2(9.8)] m =44.1 m Let h = height of stone at 4s. Using the relation, S = ut + ½ at^{2} So, h = [(29.4)(4)-1/2 (9.8) (4)^{2}]m = [117.6-78.4] m = 39.2 m From the above observation we conclude that, the height would be 39.2 m. Refer this video to know more about, “Motion Under Gravity”. Problem: A car moving in a straight line at 30 m/s slows uniformly to a speed of 10 m/s in 5 sec. Determine: (a)the acceleration of the car, (b)displacement in the third second. Solution: Let us take the direction of motion to be the +x direction. (a) For the 5 s interval, we have t = 5 s, u = 30 m/s, v = 10 m/s Using v = u + at, we have a = (v-u)/t = (10-30)/5 = -4 m/s^{2} (b) s = (displacement in 3s) - (displacement in 2s) Here u = 30 m/s, a = -4 m/s^{2}, t_{2} = 2 s; t_{3} = 3 s s = 30(3-2) + 1/2(-4)(3^{2}-2^{2}) = 20 m Alternatively S_{n} = u + a/2(2n - 1) Hence n = 3, Therefore, s^{3} = 30 + ((-4))/2 [(3)- 1]= 20 m Problem: A bullet fired into a fixed target loses half of tis velocity after penetrating 3 cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion. Solution: Let initial velocity of the bullet = u After penetrating 3 cm its velocity becomes = u/2 From v^{2} = u^{2} - 2as (u/2)^{2} = u^{2} - 2as Thus, a = u^{2}/8 Let further it will penetrate through distance x and stops at some point. 0 = (u/2)^{2} - 2 (u^{2}/8)x Therefore, x = 1 cm Problem: An anti-aircraft shell is fired vertically upwards with a muzzle velocity of 294 m/s. Calculate (a) the maximum height reached by it, (b) time taken to reach this height, (c) the velocities at the ends of 20^{th} and 40^{th} second. (d) When will its height be 2450 m? Given g = 9.8 m/s^{2}. Solution: (a) Here, the initial velocity u = 294 m/s and g = 9.8 m/s^{2} Thus, the maximum height reached by the shell is, H = u^{s}/2g = 294^{2}/(2 x 9.8) = 4410 m = 4.41 km (b) The time taken to reach the height is, T = u/g = 294/9.8 = 30 s (c) The velocity at the end of 20^{th} second is given by v = u - gt = 294 - 9.8 x 20 = 98 m/s upward, and the velocity at the end of 40^{th} second is given by, v = 294 - 9.8 x 40 = -98 m/s The negative sign implies that the shell is falling downward. (d) From the equation, H = ut + (1/2)gt^{2} or 2450 = 2941 t - 1/2 x 9.8 t^{2} or t^{2} - 60 t + 500 = 0 Therefore, t = 10 s and 50 s. At t = 10 s, the shell is at a height of 2450 m and is ascending, and at the end of 50 s it is at the same height, but is falling.
Whenever a body is released frrom a height, it travels vertically downward towards the surface of earth.This is due to the force of gravitational attraction exerted on body by the eareth. The acceleration produced by this force is called acceleration due to gravity and is denoted by ‘g’. Value of ‘g’ on the surface of earth is taken to the 9.8 m/s^{2} and it is same for all the bodies. It means all bodies (whether an iron ball or a piece of paper), when dropped (u=0) from same height shoulfd fall with same rapidity and should take same time to reach the earth. Our daily observation is contrary to this concept. We find that iron ball falls more rapidlly than piece of paper. This is due to the presence of air which offers different resistance to them. In the absence of air both would have taken same time to reach the surface of earth.
When a body is dropped from some height (earth's radius = 6400 km), it falls freely under gravity with constant acceleration g (= 9.8 m/s^{2}) provided the air resistance is negligible small. The same set of three equations of kinematics (where the acceleration remains constant) are used in solving such motion. Here, we replace by and choose the direction of y-axis conveniently. When the y-axis is chosen positive along vertically downward direction, we take as positive and use the equation as
v = u + gt, v^{2} = u^{2} + 2gh, and h = ut + 1/2gt^{2}
where h is the displacement of the body and u is initial velocity of projection in the vertically downward direction. However, if an object is projected vertically upward with initial velocity u, we can take y-axis positive in the vertically upward direction and the set of equations reduces to
v = u - gt, v^{2} = u^{2} - 2gh, and h = ut - 1/2gt^{2}
In order to avoid confusion in selecting as positive or negative, it is advisable to take the y-axis as positive along vertically upward direction and point of projection as the origin. We can now write the set of three equations in the vector form:
,
and
where h is the displacement of the body.
The motion of a particle is described by the equation u = at. The distance travelled by the particle in the first 4 second.
Because for the motion u = at. So acceleration is uniform which is equal to a.
Therefore, Distance traveled = 1/2[(a)(4)^{2}] = 8a
A body moving with a constant retardation in straight line travels 5.7 m and 3.9 m in the 6th and 9th second, respectively. When will the body come momentarily to rest?
A body moving with initial velocity u and acceleration a, traverses distance S_{n} in n^{th}second of its motion.
S_{n} = u + (1/2)(2n - 1)a => 5.7 = u + (1/2)(2 x 6 - 1) a
or 5.7 = u + (11/2) a
and 3.9 = u + (1/2)(2 x 9 - 1)a or, 3.9 = u + (17/2) a
Solving eqns. (1) and (2) we get, u = 9 m/s and a = -0.6 m/s^{2}.
If the body stops moving after t seconds, then from the relation v=u+at
Thus, 0 = 9 + (-0.6)t or, t = (9/0.6)s = 15s
A stone, thrown up is caught by the thrower after 6s. How high did it go and where was it 4 s after start? g = 9.8 m/s^{2}
Time to go up and come back = 6s
Thus, time to reach the highest point = (6/2) s = 3s
From point of projection to the highest point we have
u = ?, v = 0, a = -9.8 m/s^{2}, t = 3s
Using the relation, v = u + at
0= u – 9.83
Thus, u = 29.4 m/s^{2}
Maximum height, H = u^{2}/2g = [(29.4)^{2}/2(9.8)] m =44.1 m
Let h = height of stone at 4s.
Using the relation, S = ut + ½ at^{2}
So, h = [(29.4)(4)-1/2 (9.8) (4)^{2}]m
= [117.6-78.4] m = 39.2 m
From the above observation we conclude that, the height would be 39.2 m.
A car moving in a straight line at 30 m/s slows uniformly to a speed of 10 m/s in 5 sec. Determine:
(a)the acceleration of the car,
(b)displacement in the third second.
Let us take the direction of motion to be the +x direction.
(a) For the 5 s interval, we have t = 5 s, u = 30 m/s, v = 10 m/s
Using v = u + at, we have
a = (v-u)/t = (10-30)/5 = -4 m/s^{2}
(b) s = (displacement in 3s) - (displacement in 2s)
Here u = 30 m/s, a = -4 m/s^{2}, t_{2} = 2 s; t_{3} = 3 s
s = 30(3-2) + 1/2(-4)(3^{2}-2^{2}) = 20 m
Alternatively
S_{n} = u + a/2(2n - 1)
Hence n = 3,
Therefore, s^{3} = 30 + ((-4))/2 [(3)- 1]= 20 m
A bullet fired into a fixed target loses half of tis velocity after penetrating 3 cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion.
Let initial velocity of the bullet = u
After penetrating 3 cm its velocity becomes = u/2
From v^{2} = u^{2} - 2as
(u/2)^{2} = u^{2} - 2as
Thus, a = u^{2}/8
Let further it will penetrate through distance x and stops at some point.
0 = (u/2)^{2} - 2 (u^{2}/8)x
Therefore, x = 1 cm
An anti-aircraft shell is fired vertically upwards with a muzzle velocity of 294 m/s. Calculate (a) the maximum height reached by it, (b) time taken to reach this height, (c) the velocities at the ends of 20^{th} and 40^{th} second. (d) When will its height be 2450 m? Given g = 9.8 m/s^{2}.
(a) Here, the initial velocity u = 294 m/s and g = 9.8 m/s^{2}
Thus, the maximum height reached by the shell is,
H = u^{s}/2g = 294^{2}/(2 x 9.8) = 4410 m = 4.41 km
(b) The time taken to reach the height is, T = u/g = 294/9.8 = 30 s
(c) The velocity at the end of 20^{th} second is given by
v = u - gt = 294 - 9.8 x 20 = 98 m/s upward,
and the velocity at the end of 40^{th} second is given by,
v = 294 - 9.8 x 40 = -98 m/s
The negative sign implies that the shell is falling downward.
(d) From the equation,
H = ut + (1/2)gt^{2} or 2450 = 2941 t - 1/2 x 9.8 t^{2}
or t^{2} - 60 t + 500 = 0
Therefore, t = 10 s and 50 s.
At t = 10 s, the shell is at a height of 2450 m and is ascending, and at the end of 50 s it is at the same height, but is falling.
Acceleration of all these bodies is constant.
Acceleration is always directed downward.
All bodies, when dropped (u=0) from same height should fall with same rapidity and should take same time to reach the earth.
A stone is released with zero velocity from the top of a tower reaches the ground in 4 second, the height of the tower is about
(a) 20 m (b) 40 m
(c) 80 m (d) 16 m
A stone is thrown vertically upwards with an initial velocity of 30 m/s. The time taken for the stone to rise to its maximum height is
(a) 0.326 s (b) 3.26 s
(c) 30.6 s (d) 3.06 s
Two balls are dropped from same height at 1 second interval of time. The separation between the two balls after 3 seconds of the drop of first ball is:
(a) 50 m (b) 40 m
(c) 35 m (d) 25 m
A ball takes t second to fall from a height h_{1} and 2t seconds to fall from a height h_{2}. Then h_{1}/h_{2} is:
(a) 0.5 (b) 0.25
(c) 2 (d) 4
c
d
b
You might like to refer Introduction to Motion in One Dimension.
For getting an idea of the type of questions asked, refer the Previous Year Question Papers.
Click here to refer the most Useful Books of Physics.
To get answer to any question related to motion under gravity click here.
To read more, Buy study materials of Kinematics comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Physics here.
Signing up with Facebook allows you to connect with friends and classmates already using askIItians. It’s an easier way as well. “Relax, we won’t flood your facebook news feed!”
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
Relative Motion Relative Velocity In spite of our...
Solved Examples on Mechanics:- Example 1 :- A...
Introduction to Motion in One Dimension A body:- A...
Motion in Two Dimensions In this part, we discuss...
Circular Motion Now we shall discuss another...
Motion in a Straight Line with Acceleration...
Graphical Representation and Equations of Motion...
Motion of Projectile Now we discuss some example...