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Two balls $A$ and $B$ are thrown with same velocity $u$ from the top of a tower. Ball $A$ is thrown vertically upwards and the ball $B$ is thrown vertically downwards. If $t_A = 6\ s$ and $t_B = 2\ s$, then the height of the tower is ......... $m$
$80$
$60$
$45$
none of these
Solution
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 is $0$ equal to the time taken by ball to reach the initial place and it further takes 2 secto reach the ground because acceleration and retardation is same. $\left(10 \frac{\mathrm{m}}{\mathrm{sec}^{2}}\right)$ Hence $u=0$
$a=\frac{v-u}{t}$
$v=20 \frac{\mathrm{m}}{\mathrm{sec}}$
For the second ball the ball is thrown with initial speed $20 \frac{\sin }{\sec }$ and $t=2 \sec$ $S=u t+\frac{1}{2} a t^{2}$
By substituting the given values in above equation we get $S=60 \mathrm{m}$
Hence, the height of the tower is $60 \mathrm{m}$
