Question

A ball is thrown up from the top of a tower at 20 m/s velocity. It took the ball 6 sec to fall to the bottom. For how long was the ball above the tower?

• A. 2 sec
• B. 4 sec
• C. 6 sec
• D. 1 sec

Answer 1

The correct option is B

4 sec

When the ball was going up till B then its displacement was increasing.

When the ball started to fall from B to C its displacement was still positive but decreasing when ball is at C its displacement from the starting position is 0.

So when the ball comes back to C after falling from B then its displacement becomes O

So, we know ball was thrown with 20 m/s up

u = + 20 m/s

s = 0

Acceleration = - 10 $m/s^2$

As always the bell is under the influence of gravity

So $a=-10m/s^2$

$s=ut+\frac{1}{2}a{t}^2$

$o=20.t+\frac{1}{2}\times (-10)t^2$

t(20 - 5t) = 0

t = 0 or 4 sec

The equation gives 2 answers for t at which s = 0

By analyzing we see that at t = 0 particle was at starting point only so displacement = 0

At, t = 4 s = 0

Means ball comes back to starting point after its journey.