# 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?

Dec 14, 2015

$27 m$

#### Explanation:

The initial velocity of the ball in the vertical direction is
${u}_{y} = 46 \sin {30}^{\circ} = 23 m / s$.

It accelerates downwards under gravity at a constant rate of $9 , 8 m / {s}^{2}$ 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 :

${v}^{2} = {u}^{2} + 2 a x$

$\therefore x = \frac{{v}^{2} - {u}^{2}}{2 a} = \frac{{0}^{2} - {23}^{2}}{2 \times - 9 , 8} = 27 m$.