Question

A ball is dropped from the top of a tower of
height 80 m. At the same time, another ball is
projected horizontally from the tower. Find the
time taken by both the balls to reach the ground.
(Take g = 10 m s-2)​

Answer 1

Explanation:

Let us take the top of the tower to be the origin of the coordinates system and the uoward direction as positive and down direction as negative. Let the vertical direction be the Y-axis of the coordinates system and the horizontal direction be the X-axis.

We can get the time taken by the ball dropped from top of a tower 80 m high using the relation; s = u t + ½ g t². Here u = initial velocity with which the ball is dropped= 0 m/s as the ball is just released; t = time taken to cover s , g = acceleration due to gravity = -10 m/s² and s = displacement = - 80m

-80 = 0 × t + ½ ×(- 10) t²; ==> -80 = -5 t²; or t² = 16 or t =± 4 s. t = -4s is meaningless. So the correct solution is t = 4 s.

So the ball dropped will reach the ground in 4 s.

The ball which is projected horizontally will also take 4 s to reach the ground. Since the ball is projected with a horizontal velocity, the component of velocity of projection along the vertical direction is zero (because the component of velocity along the vertical direction would be v Cos 90°= v×0 = 0). So its vertical motion would be similar as that of the first ball and it would also take 4 s to reach the ground.

HOPE ITS GONNA HELPFUL FOR Y'LL

Answer 2

Answer:

4 second

Explanation:

case of ball 1 :

s = 80 m

u = 0 m/s

g = 10 m/s^2

t = ?

therefore ,

s = ut + 1/2 gt^2

80 = 0 × t + 1/2 ×10 ×t^2

80 = 5 t^2

t^2 = 16

t = 4 second

so, the time taken by the ball 1 is 4 second

case of ball 2:

second ball also take the same time 4 second

because, the vertically velocity is 0 v  = v cosФ  

u = v(cos 90) = v×0 = 0m/s

so, the second ball take the same time due to same vertical velocity