Question

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

• A. $52$
• B. $26$
  
• C. $51$
  
• D. $20$
  

Answer 1

Answer: (A)

$52$

Here the time 5 sec is for the body to rise from 10.2 m elevation to the maximum height and another 5 sec is for fall back to 10.2 m elevation.

Since at highest point the velocity of the body is zero, so the velocity at 10.2 m height is $v_h=gt=10\left(5\right)=50m/s$ (using $v=u+at$ here $v=0,u=v_h,a=-g$)

Using conservation of energy, $mgh=\frac{1}{2}m[v_0^2-v_h^2]$

or $v_0^2=2gh+{50}^2=2\left(10\right)\left(10.2\right)+{50}^2$

or $v_0=52m/s$