Question

A ball is thrown vertically upwards with a speed of $10$ m/s from the top of the tower $400$ m high and another is thrown vertically down wards with the same speed simultaneously. The time difference between them in reaching the ground is ( $g=10m/s^2$ )

• A. 12s
• B. 7s
• C. 2s
• D. 1s

Answer 1

The correct option is C 2s

Time taken by $B:$

$\Rightarrow 400=10\left(t\right)+\frac{1}{2}g{t}^2$
$\Rightarrow 400=10t+5t^2$
$\Rightarrow t^2+2t-80=0$
$\Rightarrow t=\frac{-2\pm \sqrt{324}}{2}=\frac{-2\pm 18}{2}=8sec$

Time taken by $A:$

$\Rightarrow 400=10\left(t\right)-\frac{1}{2}g{t}^2$
$\Rightarrow 400=10t-5t^2$
$\Rightarrow t^2-2t+80=0$
$\Rightarrow t=1sec$
Hence, time difference is $7$ sec.