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 $(g=10m/s^2)$

• A. $60m,15m/s$
• B. $80m,20m/s$
• C. $60m,20m/s$
• D. $40m,10m/s$

Answer 1

Answer: (C)

$60m,20m/s$

When the ball which was thrown upward with speed$u$comes down at the height from which it was thrown the speed becomes equal to the initial speed i.e$u$the time taken by the ball to reach upward where the speed is$$0$$equal to the time taken by ball to reach the initial place and it further takes $2sec$to reach the ground because acceleration and retardation is same.$\left(10\frac{m}{se{c}^2}\right)$

Hence

$u=0$

$a=\frac{v-u}{t}$

$v=20\frac{m}{sec}$

For the second ball the ball is thrown with initial speed $20\frac{m}{sec}$ and $t=2sec$

$S=ut+\frac{1}{2}a{t}^2$

By substituting the given values in above equation we get

$S=60m$

Hence, the height of the tower is $60m$