Question

a cricket ball is thrown up with a speed of 19.6 m/s. the maximum height it can reach be?

Answer 1

Here we should use newton's second eqn. of motion 
that is= "V^2-U^2=2GH" 

So here u= 19.6m/s. and v=0m/s as after reaching a certain height v becomes 0.

now the eqn. changes to -u^2=2gh which can be written as h=-u^2/2g
And here even g is also neagative as the ball is thrown upwards

so, h= -384.16/2*-9.8
    h= 19.6m

Answer 2

Answer:

The answer is 19.6m

Explanation:

This is a very easy problem. Here we need the maximum height.First of all remember that this is a projectile motion. So the maximum height will be given by u^2/2g...thats it