Question

A ball is short form the ground into the air at the height 9.1m its velocity to be abserved to be v=7.6i+6.1jm/s to what maximum height the ball will rise

Answer 1

The Initial velocity of the ball be (u1 i + u2 j).  

Vertically:  

6.8^2 = u2^2 - 2g * 9.1  

u2^2 = 6.8^2 + 2g * 9.1  

u2 = 15.0 m/s.  

Horizontally there is no change in velocity:  

u1 = 7.2 m/s ...(1)  

At the maximum height h m, there is no vertical velocity.  

0 = u2^2 - 2gh ...(2)  

h = u2^2 / 2g  

h = 11.5m.