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A body is projected vertically upwards. If t1 and t2 be the times at which it is at height h above the point of projection while ascending and descending respectively, then h is, (1)1/2gt1t2 (2)gt1t2 (3)2g(t1+t2) (4)4g(t1+t2)

A body is projected vertically upwards. If t1 and t2 be the times at which it is at height h above the point of projection while ascending and descending respectively, then h is, (1)1/2gt1t2 (2)gt1t2 (3)2g(t1+t2)  (4)4g(t1+t2)

Grade:12

6 Answers

Croma Tab
21 Points
6 years ago
t1 is the time the particle takes to reach h.
so,
              h = ut1 - ½ gt12           \Rightarrow         u = (h + ½ gt12) / t1      (equation 1)
 
t2 is the time taken by the particle to reach the same point again.
so,
             h = ut2 - ½ gt22
 
   substituting 1 in the above equation:
 
            h = (h + ½ gt12) t2/t1 – ½ gt22
 
   solving
 
            h = ½ gt1t2
 
 
nithishagattu
20 Points
6 years ago
Let t1 be the time where particle reach height " h" and let t2 be the time where particle attains same height "h" again. 
Time of flight T=t1 +t2
  • h=(usinθt)+1/2gt2​ ; u is velocity of projection. 
gt2-2usinθt+2h=0
the roots of this equation are t1and t2
t1t2=2h/g, t1+t2=2usinθ/g [for ax2+bx+c=0,sum of the roots=-b/a and product of the roots=c/a] 
By solving these equations we get h=1/2gt1t2
​
Aakash chalise
14 Points
5 years ago
a body is projected vertically upwards from the ground.It is found at the same elevation at t=3 and t=7s after projection.Find the projection speed.
Aakash chalise
14 Points
5 years ago
a body is projected vertically upwards from the ground.It is found at the same elevation at t=3 and t=7s after projection.Find the projection speed.
Answer.
U=g÷2(t1+t2)
  =5×(3+7)
  =5×10
  =50m/s.
 
U=g÷2(t1+t2)
 =5×(3+7)
 =5×10
 =50
Pranav Kumar Varshney
15 Points
4 years ago
u = velocity of projection (upward).
a = acceleration (downward) = - g
s = displacement after time t = h
Applying the relation s = u t + (1/2) a t², we have
h = u t - (1/2) g t²
=> 2h = 2ut - gt²
=> gt² - 2ut + 2h = 0 .
This is a quadratic equation in t which shows that for a given h there are two values of t.
               Solving the equation for t using Quadratic formula(x=(-b± √b2-4ac)/2a)
t=(2u± √4u2-8gh)/2g
t=(u± √u2-2gh)/g
=> t1 = {u + √(u²- 2gh)} / g and t2 = {u - √(u²- 2gh)} / g
Multiplying, (t1)(t2) = {u² - (u² - 2gh)} / g² = 2gh / g²
=>(t1)(t2) = 2h / g
=> h = (1/2)g(t1)(t2) 
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the answer to your problem below.
 
we have,
u = velocity of projection (upward).
a = acceleration (downward) = – g
s = displacement after time t = h
Applying the relation s = ut + (1/2)at², we have
h = ut – (1/2)gt²
or, gt2 – 2ut + 2h = 0
Now, since it is a quadratic equation, and its roots are t1 and t2
Hence, Product of roots = c/a = 2h/g
or, t1t2 = 2h/g
Hence, h = (1/2)gt1t2
 
Hope it helps.
Thanks and regards,
Kushagra

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