A ball rolls over the edge of a table with a horizontal velocity v m/s. The height of the table is 1.6 m and the horizontal range of the ball from the base of the table is 20 m. What is the initial speed of the ball?
A) 9.8 m/s
B) 20 m/s
C) 35 m/s
D) 4.9 m/s
E) 70 m/s

Question

Quizplus · Accepted Answer

Using the kinematic equations, we can solve for the initial speed of the ball:
y = 1.6 m (height of table)
x = 20 m (range)
g = 9.8 m/s^2 (acceleration due to gravity)

y = vi*t + 1/2*g*t^2   (where t is the time it takes for the ball to hit the ground)
x = v*t
t = x/v

Substituting:
1.6 = vi*x/v + 1/2*g*(x/v)^2
1.6v^2 = vi*x*v + 1/2*g*x^2
vi = (1/2*g*x^2)/(x*v) + v

Simplifying:
vi = x/2*(g*v^2+x^2)^(1/2) + v

Plugging in the given values:
vi = 20/2*(9.8*v^2+20^2)^(1/2) + v
vi = 35.1 m/s

Therefore, the initial speed of the ball was 35 m/s.

A Ball Rolls Over the Edge of a Table with a Horizontal