Question

A ball is thrown vertically upwards from the ground and a student gazing out the window sees moving upward past him at $10\text{ m}{{\text{s}}^{-1}}$. The window is at 15 m above the ground level. The velocity of the ball 3 seconds after it was projected from the ground is? (take g = $10\text{ m}{{\text{s}}^{-1}}$)
A. 10 m/s, up
B. 20 m/s, up
C. 20 m/s, down
D. 10 m/s, down

Answer 1

Hint: When an object is thrown upward with an initial velocity u, the acceleration due to gravity acts against the motion of the ball decreasing its velocity till the velocity of the ball reaches zero while in flight.

Complete step-by-step answer:
It is mentioned in the question that a student who is at a height of 15 m from the ground sees the velocity of the ball as $10\text{ m}{{\text{s}}^{-1}}$. So let this velocity be taken as v and the height as h. So according to the newton’s third equation of motion, ${{\text{v}}^{\text{2}}}-{{\text{u}}^{\text{2}}}=-2\text{gh}$. The value of g is negative since the acceleration due to gravity is acting opposite to the motion of the body. So substituting the values in the equation, we get,
$\begin{align}
  & {{\left( 10 \right)}^{2}}-{{u}^{2}}=-2\times (10)\times (15) \\
 & \Rightarrow {{u}^{2}}=300+100=400 \\
\end{align}$
$\therefore \text{u}=20\text{ m/s}$
So the initial velocity with which the ball was projected was 20 m/s.
The velocity of the ball after 3 seconds into the flight can be found out using the equation,
$\text{v}=\text{u}+\text{at}$
Here u=20 m/s, a = -g, t=3 seconds.
$v=20+\left( -10 \right)\times 3$
$\therefore \text{ v}=-10\text{ m/s}$
Minus sign indicates the downward direction. So at t=3 seconds, the ball is coming down at a velocity of 10 m/s.
So the answer to the question is option (D) 10 m/s, down.

Note: During freefall, gravity is the only force acting on the body.Time to reach the maximum height in this problem is 2 seconds.So after the 3 seconds, the ball has already reached its maximum height and has proceeded to free fall.