Question

After falling from rest from a height of 30 m, a 0.50-kg ball re-bounces upward, reaching a height of 20 m. If the contact between the ball and the ground lasted 2.0 ms, what average force was exerted on the ball?

Answer 1

do you have the answers? the answer i got is -6060N

Answer 2

Yes I do have the answer. It is 11,000N but I need a thorough explanation of how to arrive at it.

Answer 3

First, the difference on the heights between bounces is due to energy lost as heat and sound in the collition on the floor. These can be considered as work done by the floor. You see, work has these two definition which we can use to find the average force.\[W=\Delta E=Fd \rightarrow F=\frac{ \Delta E }{ d }=\frac{ E _{f}-E _{i} }{ t }=\frac{ mgh _{f}-mgh _{i} }{ t }=\frac{ mg(h _{f}-h _{i}) }{ t }\]Therefore:\[F=\frac{ 0.5kg9.8\frac{ m }{ s ^{2} }(20m-30m) }{ 2ms \frac{ 1s }{ 1000ms } }=-24500N\]Thats the answer.

Answer 4

I like your approach to the question but the book says 11,000N

Answer 5

using v^2 = u^2 + 2ax {this follows from v = u + at and x = ut + 1/2(a)t^2) - it just eliminates t from the story}
so, to calc velocity of ball as it hits ground:
v^2 = 2 * 10 * 30
v = sqrt(600) ~ 24.5m/s

using same to calc velocity ball as it leaves the ground again:

0 = u^2 + (2)(-10)(20)
u = sqrt(400) = 20m/s

F = mass * acceleration (Newton's 2nd)
the force pushing up on the ball during contact does two things:
1) it accelerates the ball back in the direction it came from
2) it counteracts gravity

ball acceleration = dv/dt where dv is the change in the ball's velocity and dt is the contact time

dv = 24.5 - (-20) = 44.5m/s [remember velocity is a vecter, it has quantity and direction]

dt = 2ms (the contact time during which the ball is accelerated back up)

F = m(a + g) = m *( dv/dt +g) = 0.5 * {(44.5 / (0.002)) +10} = 11,130N


close enough?

Answer 6

I think Irishboy is right. I just saw that I irracionally changed "d" for "t" in the middle of the equation. Wow that was a very creative approach.