Question

A bowling ball falls freely from rest for 10 seconds before it hits the ground. Neglect air resistance.Is the average speed of the ball during the first 5 seconds greater than, less than, or the same as the average speed of the ball during the last 5 seconds?

Answer 1

100 m/s.

Step-by-step explanation:

Reasoning: Speed increases at a rate of 10 m/s (actually 9.8 m/s) every second. Thus after 10 seconds, the speed is 10 x 10 = 100 m/s.

Answer 2

the average speed of the ball during the first 5 seconds is less than the average speed of the ball during the last 5 seconds

Explanation:

The average speed is defined as

`v_{\text{avg}}=(v_(i1)+v_(f1))/(2)`

where v_i = inital speed and v_f = final speed.

Now for the first five seconds, we know that the ball started from rest; therefore, v_i = 0, and so the average speed is

`v_{\text{avg}}=(0+v_(f1))/(2)`
`\boxed{v_{\text{avg}}=(v_(f1))/(2)}`

where v_f = final velocity at the end of the first 5 seconds.

Now the average velocity in the last 5 seconds is

`v_{\text{avg}}=(v_(i2)+v_(f2))/(2)`

Realising that v_i2 = v_f1 ( the initial velocity at the beginning of last 5 seconds = final velocity at the end of first 5 seconds) and that vf_2 > v_i2 gives us