Question

A body thrown vertically up from the ground passes the height $10.2m$ twice at an interval of $10s$. What was its initial velocity? (in m/s)

• A. $52$
  
• B. $53$
  
• C. $51$
  
• D. $49$
  

Answer 1

The correct option is A $52$

if the body moving vertically passes twice in 10 sec then it is clear that it will go up for 5 sec and next 5 sec will be consumed by body to come down back.

distance traveled in 5 sec starting from rest and with acceleration 10(returning from top using $S=ut+\frac{1}{2}a{t}^2$), is 125 meters. so total height is $(125+10.2=135.2)$

initial velocity is calculated by using $v^2-u^2=2aS=>v=\sqrt{2704}=52m/s.$