Question

A physics student stands on a cliff overlooking a lake and decides to throw a golf ball to her friends in the water below. She throws the golf ball with a velocity of 19.5 m/s at an angle of 33.5 ∘ above the horizontal. When the golf ball leaves her hand, it is 14.5 m above the water. How far does the golf ball travel horizontally before it hits the water? Neglect any effects of air resistance when calculating the answer.

Answer 1

Answer:

50.5 m

Explanation:

We are given that

$\theta=33.5^{\circ}$

Velocity=v=19.5m

Vertical component of initial velocity=$v_y=vsin\theta=19.5sin33.5=10.8m/s$

Height=h=-14.5 m

Acceleration due to gravity=$g=-9.8m/s^2$

$s=v_yt+\frac{1}{2}gt^2$

$-14.5=10.8t+\frac{1}{2}(-9.8)t^2$

$4.9t^2-10.8t-14.5=0$

By solving we get

$t=-0.9, t=3.1$

Time cannot be negative

Therefore, time=3.1 s

Horizontal component of velocity=$v_x=19.5cos33.5=16.3m/s$

Distance=$x=v_xt=16.3\times 3.1=50.5 m$

Hence, the golf ball travel horizontally before it hits the water=50.5 m