Answer:
50.5 m
Explanation:
We are given that
[tex]\theta=33.5^{\circ}[/tex]
Velocity=v=19.5m
Vertical component of initial velocity=[tex]v_y=vsin\theta=19.5sin33.5=10.8m/s[/tex]
Height=h=-14.5 m
Acceleration due to gravity=[tex]g=-9.8m/s^2[/tex]
[tex]s=v_yt+\frac{1}{2}gt^2[/tex]
[tex]-14.5=10.8t+\frac{1}{2}(-9.8)t^2[/tex]
[tex]4.9t^2-10.8t-14.5=0[/tex]
By solving we get
[tex]t=-0.9, t=3.1 [/tex]
Time cannot be negative
Therefore, time=3.1 s
Horizontal component of velocity=[tex]v_x=19.5cos33.5=16.3m/s[/tex]
Distance=[tex]x=v_xt=16.3\times 3.1=50.5 m[/tex]
Hence, the golf ball travel horizontally before it hits the water=50.5 m
