Question

As.A cricket ball is thrown up with a speed of 19.6 ms. The
maximum height it can reach is
(A) 9.8 m
(B) 19.6 m
(C) 29.4 m
(D) 39.2 m​

Answer 1

Answer:

Explanation:

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