Discussion Forum

 A ball is projected vertically up from ground. Boy A standing at the window of first floor of a nearby building observes that the time interval  between the ball crossing him while going up and the ball crossing him while going down is 2s . Another boy B  standing  on the second  floor notices that time interval between the ball passing him twice, during up motion  and down  motion is 1s . Calculate the difference between  the vertical positions  of boy B and boy A (Assume , acceleration  due to gravity ,g=10 m/s²)

Answer:

Explanation:

 The figure given below , shows the vertical motion of ball,

Here, ball crosses the 1st floor in 2s, so it goes up in 1s and goes down in next one second.

So,   $v-v_{A}=-gt$

 $\Rightarrow$    0-$v_{A}=-10(1) \Rightarrow   v_{A}=10m/s$

$\because$ Boy, B observed the ball crossing him while up and down motion in 1s, So , time taken by the ball to reached the floor B from A is 0.5s

So,    s=$v_{A}t- \frac{1}{2} gt^{2}$

=$10 \times 0.5 -\frac{1}{2}\times 10 \times (0.5)^{2}$

 =5-1.25=3.75 m

 Hence, the correct option is (b)