Question

A ball is projected upwards from the foot of a tower. The ball crosses the top of the tower twice after an interval of $6s$ and the ball reaches the ground after $12s$. The height of the tower is $(g=10m/s^2)$

• A. $120m$
• B. $135m$
• C. $175m$
• D. $80m$

Answer 1

The correct option is B $135m$


According to question,
$t_{BC}=\frac{6}{2}=3s$
$t_{AC}=\frac{12}{2}=6s$
$\therefore t_{AB}=3s$
By using first equation of motion; as at the top $(v=0)$
$\therefore 0=u-\left(10\right)6$
or $u=60m/s$
When we throw the ball upward then the distance travelled by ball in initial 3 second is equal to the height to tower
i.e., $h=u{t}_{AB}-\frac{1}{2}g{t}_{AB}^2$
$=\left(60\right)\left(3\right)-\frac{1}{2}\left(10\right)(3{)}^2$
$=135m$