Question

A boy standing on a long railroad car throws a ball straight upwards with a speed of $10\text{m/s}$ at $t=0$ when the car starts moving on a horizontal road with an acceleration of $1{\text{m/s}}^2$. How far behind the boy will the ball fall on the car? (Take $g=10{\text{m/s}}^2$).

• A. $4\text{m}$
• B. $2\text{m}$
• C. $4.5\text{m}$
• D. $8\text{m}$

Answer 1

The correct option is B $2\text{m}$

Whenever an object is thrown from a moving frame, the velocity of the frame is transfered to the object.
Hence, for ball:
(considering vertically upwards as $+ve$ direction)
$u_y=+10\text{m/s}$
$u_x=0$
$a_y=-10{\text{m/s}}^2$
$a_x=0$
Time taken to return to the car,
$T=\frac{2u_y}{g}=\frac{20}{10}$
$\therefore T=2\text{s}$
For, car initial velocity $u_x=0$
$a_x=1{\text{m/s}}^2$
Horizontal distance travelled by car during the time for ball to return,
$X_1=\frac{1}{2}{a}_x\times T^2$
$\Rightarrow X_1=\frac{1}{2}\times 4=2\text{m}$
$\because$ Ball has no horizontal component of velocity, hence it falls by $2\text{m}$ behind car.