Question

A golfer hit a golf ball at 35° from the ground with a velocity of 43m/s what is the height of the golf ball at the apex of its flight

Answer 1

We can find the height of the golf ball at the apex of its flight from the kinematic equation:

$$v_y^2=v_{0y}^2-2gy_{max},$$

here, $v_y=0\ \dfrac{m}{s}$ is the velocity of the golf ball at the maximum height, $v_{0y}=v_0sin\theta$ is the vertical component of the initial velocity of the golf ball, $v_0=43\ \dfrac{m}{s}$ is the initial velocity of the golf ball, $\theta=35^{\circ}$ is the launch angle, $y_{max}$ is the height of the golf ball at the apex of its flight (or the maximum height attained by the golf ball) and $g = 9.8 \ \dfrac{m}{s^2}$ is the acceleration due to gravity.

Then, from this formula we can find $y_{max}$:

$$0=v_0^2sin^2\theta-2gy_{max},$$ $$y_{max}=\dfrac{v_0^2sin^2\theta}{2g} = \dfrac{(43\ \dfrac{m}{s})^2 \cdot sin^2 35^{\circ}}{2\cdot 9.8 \ \dfrac{m}{s^2}} = 31 \ m.$$

Answer:

$y_{max}=31 \ m.$