Question

a particle is thrown vertically upward .its velocity at half of the height is 10m/s,then the maximum height is what?g=10m/s

Answer 1

Let the velocity with which it was projected be u  

let the height achieved by the particle = h  

v = 10 m/s  

So by kinematic equation, at height h/2  

v^2 = u^2 - 2g(h/2)  

=> 10^2 = u^2 - gh --- 1.  

Again at the maximum height h, particle's velocity v = 0  

v^2 = u^2 - 2gh  

=> 0^2 = u^2 - 2gh  

=> u^2 = 2gh ----2  

plugging value of u^2 from eqn. 2 to eqn 1  

10^2 = u^2 - gh  

=> 10^2 = 2gh - gh  

=> gh = 100  

=> h = 100/g  

Taking value of g = 10 m/s^2  

h = 100/10 = 10 m