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 $6\text{s}$ and the ball reaches the ground after $12\text{s}$. The height of the tower is ($g=10m/s^2$)

• A. $120\text{m}$
• B. $135\text{m}$
• C. $175\text{m}$
• D. $80\text{m}$

Answer 1

The correct option is B $135\text{m}$


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