Grade 11Mechanics

a ball is projected upwards from the top of the tower with the velocity 50 m/s making an angle 30 degree with the horizontal.The height of tower is 70m.After how many seconds from the instant of throwing will the ball reach the ground?

maady manis

11 Years agoGrade 11

6 Answers

Sumit Majumdar

11 Years ago

Dear student,
The Total time after which the ball would reach the ground would be given by:
$T=\frac{2usin\theta}{g}+\sqrt{\frac{2h}{g}}=\frac{2\times 50\times sin30^{\circ}}{10}+\sqrt{\frac{2\times 70}{10}}=5+3.74=8.74 sec$ Regards
Sumit

Ananya Sharma

11 Years ago

But the correct answer is 7s. Pls. can u help me futher with my answer

Akshay

11 Years ago

@ananya: yes the answer is 7.

Let throwing point be (0,0). Horizontal be x-direction and vertical be y-direction
Initial velocity = (50cos30) i + (50sin30) j,

a = -10 j

s = ut + 0.5*a*t²,

s = (50*cos30*t) i + (50*sin30*t – 5*t²) j,

it will reach the ground when y-coordinate of ball is -70,

So, (50sin30*t – 5t²) = -70

or, 25*t-5*t²=-70, t²-5t-14=0

(t-7)(t+2)=0 => t=7

JohnCenaToTheRescue

8 Years ago

Why can`t we calculate it by sumit`s formula.

Mukul

8 Years ago

Taking vertical downward motion of projectile from point of projection to ground ,we have

u=-50sin30°=-25m/s

a=+10m/s^2. ;s=70m ,t=?

Using S=ut+1/2at^2

So, 70=-25×t+1/2×10×t^2

Or 5t^2-25t-70=0

Or t^2-5t-14=0

On solving,t=7s.{Here t^2 means t raise to power 2}

Rishi Sharma

5 Years ago

Dear Student,
Please find below the solution to your problem.

Let throwing point be (0,0). Horizontal be x-direction and vertical be y-direction
Initial velocity = (50cos30) i + (50sin30) j,
a = -10 j
s = ut + 0.5*a*t2,
s = (50*cos30*t) i + (50*sin30*t – 5*t2) j,
it will reach the ground when y-coordinate of ball is -70,
So, (50sin30*t – 5t2) = -70
or, 25*t-5*t2=-70
t^2-5t-14=0
(t-7)(t+2)=0
t=7

Thanks and Regards