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

Step-by-step 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
`(2usin\theta )/(g)`

`t_1=(2* 25.5* 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+(gt^2)/(2)`

`18.5=25.5sin(33.5)\cdot t+(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* t`

`R=25.5cos(33.5)* 3.85=81.86 m`