Question

A golf ball is launched at a 30 degree angle with a speed of 46 m/s.To the nearest tenth of a meter, what is its maximum height?

Answer 1

`27m`

Explanation:

The initial velocity of the ball in the vertical direction is
`u_y=46sin30^@=23m//s`.

It accelerates downwards under gravity at a constant rate of `9,8m//s^2` and at maximum height, its velocity in the y-direction is `v_y=0`.

We may use the equations of motion for constant linear acceleration in the y-direction to determine the maximum height `x` achieved as follows :

`v^2=u^2+2ax`

`therefore x=(v^2-u^2)/(2a)=(0^2-23^2)/(2xx-9,8)=27m`.