Question

A ball thrown vertically upwards with a speed of $20m{s}^{-1}$ from the top of a tower reaches the earth in $8s$. Find the height of the tower. Take $g=10m{s}^{-2}$

• A. $120m$
• B. $100m$
• C. $60m$
• D. $160m$

Answer 1

The correct option is D $160m$

Given:
Initial velocity, $u=20m{s}^{-1}$
Time taken, $t=8s$

Let the height of the tower be $h$.
Assume the upwards direction to be positive.

The magnitude of the displacement is the height of the tower. Final position (ground) is in downwards (or negative) direction with respect to the initial position (top of tower).
$\therefore s=-h$

Acceleration is in downwards direction.
$\therefore a=-g=-10m{s}^{-2}$

From the second equation of motion,
$s=ut+\frac{1}{2}a{t}^2$,
$\Rightarrow -h=ut-\frac{1}{2}{gt}^2$
$\Rightarrow -h=20\times 8-(\frac{1}{2}\times 10\times 8^2)$
$\Rightarrow h=160m.$

therefore, height of tower = $160m$