A ball is dropped from the 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 ms-2-

Question

Toppr Admin · Accepted Answer

Vertical velocity of both balls is initially is zero-So- -s - ut - -dfrac-1-2-at-2-80 - s - 0 - -dfrac-1-2-times10-times t-2-16 - t-2-160-t - 4 sec for both balls-

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 $$ms^{-2}$$).

Vertical velocity of both balls is initially is zero.
So, $$s = ut + \dfrac{1}{2}at^2$$
$$80 = s = 0 + \dfrac{1}{2}\times10\times t^2$$
$$16 = t^2$$
t = 4 sec for both balls.

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 $$ms^{-2}$$).

Vertical velocity of both balls is initially is zero.So, $$s = ut + \dfrac{1}{2}at^2$$$$80 = s = 0 + \dfrac{1}{2}\times10\times t^2$$$$16 = t^2$$ t = 4 sec for both balls.

Vertical velocity of both balls is initially is zero.
So, $$s = ut + \dfrac{1}{2}at^2$$
$$80 = s = 0 + \dfrac{1}{2}\times10\times t^2$$
$$16 = t^2$$
t = 4 sec for both balls.