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Josh rolled a bowling ball down a lane in 2.5s.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 last Tim’s hand

User Korbes
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1 Answer

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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?

User Skqr
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