1

Question

Two balls X and Y are thrown from top of tower one vertically upward and other vertically downward with same speed. If times taken by them to reach the ground are 6s and 2s respectively, then the height of the tower and initial speed of each ball are (g=10m/s2)

Open in App

Solution

The correct option is **D** 60m,20m/s

When the ball which was thrown upward with speeducomes down at the height from which it was thrown the speed becomes equal to the initial speed i.euthe time taken by the ball to reach upward where the speed is0equal to the time taken by ball to reach the initial place and it further takes 2secto reach the ground because acceleration and retardation is same.(10msec2)

Hence

u=0

a=v−ut

v=20msec

For the second ball the ball is thrown with initial speed 20msec and t=2sec

S=ut+12at2

By substituting the given values in above equation we get

S=60m

Hence, the height of the tower is 60m

3

View More

Join BYJU'S Learning Program

Join BYJU'S Learning Program