Answer:
The velocity of the ball before it hit the ground was (3.59 m/s, -3.32 m/s) in the horizontal and vertical direction
Step-by-step explanation:
The velocity of the ball can be calculated using the equation of motion:
v^2 = u^2 + 2as,
where
u = initial velocity,
v = final velocity,
a = acceleration due to gravity (9.8 m/s^2),
s = vertical height fallen (0.86 m).
Solving for u:
u = sqrt(v^2 - 2as)
We know the final velocity, v = 0 (the ball lands on the ground and stops), so
u = sqrt(2as) = sqrt(2 * 9.8 * 0.86) = 3.32 m/s.
The direction of the velocity before it hit the ground can be determined using horizontal distance traveled and time of flight.
The time of flight, t, can be found using:
t = sqrt(2s/a) = sqrt(2 * 0.86 / 9.8) = 0.39 s.
The horizontal velocity, vx, can be found using:
vx = d / t = 1.4 / 0.39 = 3.59 m/s.