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?

1 Answer
Dec 14, 2015

27m27m

Explanation:

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

It accelerates downwards under gravity at a constant rate of 9,8m//s^29,8m/s2 and at maximum height, its velocity in the y-direction is v_y=0vy=0.

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

v^2=u^2+2axv2=u2+2ax

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