Question

A physics student stands on a cliff overlooking a lake and decides to throw a tennis ball to her friends in the water below. She throws the tennis ball with a velocity of 25.5 m/s at an angle of 33.5∘ above the horizontal. When the tennis ball leaves her hand, it is 18.5 m above the water. How far does the tennis ball travel horizontally before it hits the water? Neglect any effects of air resistance when calculating the answer.

Answer 1

Answer:

Explanation:

Given

Initial velocity(u)=25.5 m/s

Projection angle$(\theta )=33.5^{\circ}$

ball is 18.5 m above the ground

Time of flight for zero vertical displacement is $\frac{2usin\theta }{g}$

$t_1=\frac{2\times 25.5\times sin(33.5)}{9.81}=2.87 s$

its vertical velocity will be equal to initial vertical velocity but it will be downward direction at the completion of zero vertical displacement

Now time taken to cover 18.5 m vertical displacement

$s=ut+\frac{gt^2}{2}$

$18.5=25.5sin(33.5)\cdot t+\frac{9.81\cdot t^2}{2}$

$9.81t^2+28.15t-37=0$

t=0.979 s

Thus $t_2=0.979 s$

Total time to reach ground $=t_1+t_2=2.87+0.979=3.85 s$

Horizontal Distance traveled in 3.85 s

$R=u_x\times t$

$R=25.5cos(33.5)\times 3.85=81.86 m$