Question

A tennis ball rolls off the edge of a tabletop 0.850 m above the floor and strikes the floor at a point 1.00 m horizontally from the edge of the table.

Find the time of flight of the ball.

Find the magnitude of the initial velocity of the ball.

Find the magnitude of the velocity of the ball just before it strikes the floor.

Find the direction of the velocity of the ball just before it strikes the floor.

Answer 1

given

h = 0.850 m

strikes the floor at a point 1.00 m horizontally from the edge of the table

using equations of motions  

s_y = ut +0.5*gt^2

0.850 = 0*t+0.5*9.8*t^2

t = 0.4165 s

time of flight is t = 0.4165

magnitude of initial velocity is V = s_x/t = 1.00/0.4165 m/s = 2.4 m/s

velocity of the ball just before it strikes the floor is

first vertical velocity is v_y = g*t = 9.8*0.4165 m/s = 4.0817 m/s

now the velocity magnitude is v = sqrt(V_x^2+V_y^2)

V = sqrt(2.4^2+4.0817^2) m/s = 4.735 m/s

now the direction is  

theta = arc tan (v_y/v_x)

theta = arc tan (4.0817/2.4)

theta = 59.45 degrees