Question

Josh rolled a bowling ball down a lane in 2.5 seconds. The ball travelled at a constant acceleration of 1.8 m/s/s 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 Josh’s hand?

Answer 1

Ok, so slope of the lane gives the ball a constant acceleration of 1.8 m/s for 2.5 s
[lane slope = arcsine (1.8/9.8) = 10.6°]
which increases speed of the ball by
1.8(2.5) = 4.5 m/s
The ball reaches the pins with a speed = 7.6 m/s, so it must have left Josh's hand with a speed of
7.6 - 4.5 = 3.1 m/s ANS

Answer 2

Answer:

• The ball had a speed of $3.1m/s$ when it left Josh’s hand

Explanation:

The final velocity of the ball is

$V_f = V_i + at$

The initial velocity of the ball is

$V_i = V_f - at\\\\V_i = 7.6 - (1.8*2.5)\\\\V_i = 3.1m/s$

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