Question

A ball is projected upwards from top of a tower with a velocity of 50 m/s making an angle 30° with the horizontal. The height of tower
is 70 m. After how much time from the instant of throwing will the ball reach the ground?
2

Answer 1

Explanation:

The component of velocity in the vertical-upward direction is,

v=50sin(30°)=50×0.5=25m/s

Let the time taken to reach ground from initial position be t sec

The acceleration due to gravity is, g=10m/s^2 , in the vertical-downward direction.

and the distance traveled is, h=70m, in the vertical-downward direction.

So here, h=−vt+ 1/2 gt^2

⟹70=−25t+(0.5×10×t^2)

⟹t^2−5t−14=0

Solving above quadratic equation, we get t=(−2),7

As time is always positive, t=7s

so,

After 7s from the instant of throwing will the ball reach the ground.