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 19.5 m/s at an angle of 36.5∘ above the horizontal. When the tennis ball leaves her hand, it is 15.5 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.

Explanation:

The horizontal component of the velocity = 19.5 cos 36.5 = 15.67m /s

The vertical component = 19.5 sin36.5 = 11.6 m /s

height h = 15.5

we shall first find the time of fall of the ball into water

vertical displacement = height = 15.5 ( positive and g will be positive as it is downwards .

s = ut + 1/2 gt²

15.5 = - 19.5 sin36.5  t + 1/2 x 9.8 x t²

4.9 t² - 11.6 t - 15.5 = 0

t = 3.31 s

During this period ball will travel in horizontal velocity of 15.67 m /s uniformly

horizontal displacement before touching water

= 3.3 x 15.67

= 51.7 m .