The bowling ball was going 3.1 m/s when it left Josh's hand.
We can use the following equation to relate the final velocity, initial velocity, acceleration, and time:
vf = vi + at
where:
vf is the final velocity (7.6 m/s)
vi is the initial velocity (unknown)
a is the acceleration (1.8 m/s²)
t is the time (2.5 s)
Rearranging the equation to solve for vi:
vi = vf - at
vi = 7.6 m/s - 1.8 m/s² * 2.5 s
vi = 7.6 m/s - 4.5 m/s
vi = 3.1 m/s
Therefore, the bowling ball was going 3.1 m/s when it left Josh's hand.
Question
Josh rolled a bowling ball down a lane in 2.5 s. The ball traveled at a constant acceleration of 1.8 m/s2 down the lane and was traveling at a speed of 7.6 m/s by the time it reached the pins at the end of the lane. How fast was the ball going when it left Tim's hand?