Question

Suri rolled a bowling ball down a lane in 2.5 s. The ball traveled

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 at the
beginning, when it left Suri's hand?
​

Answer 1

3.1 m/s

Step-by-step explanation:

The acceleration of the ball is given by:

`a=(v-u)/(t)`

where

v is the final velocity

u is the initial velocity

t is the time elapsed

Here we know:

a = 1.8 m/s^2 is the acceleration

v = 7.6 m/s is the final velocity

t = 2.5 is the time elapsed

Solving for u, we find the initial velocity of the ball:

`u=v-at=7.6 m/s-(1.8 m/s^2)(2.5 s)=3.1 m/s`