To test the quality of a tenis ball, you drop it onto the floor from a height of 4.00m. It rebounds to a?

Question

height of 1.65m. suppose the ball is in contact with the floor for 15.0ms. What is the magnitude of its average acceleration during the contact with the floor?

Answer 1

We need to find the change in velocity, over time, to determine the acceleration.

The ball loses energy as it is in contact with the ground, supposedly. It hits the ground with a certain velocity, and then leaves the ground with a much smaller velocity - not enough to return it to its original height.

Use motion equations to find the velocity at the instant before and after touching the ground. I'll refer to them as v1 and v2.
(v1)^2 = 2aΔ(x1) = 2 * -9.8 * 4 = -78.4; sqrt = -8.854
(v2)^2 = 2aΔ(x2) = 2 * -9.8 * -1.65 = 32.34; sqrt = 5.687

The ball changed more than the difference in these two - look at the signs. (The second one is positive because of the Δx, or change in distance, which is the opposite direction)
The ball went from -8.854 m/s to 0 m/s, and then started gaining velocity in the positive direction. That means the Δv was really 8.854+5.687 = 14.5412 m/s

Now, v = a*t solving for a = v/t = 14.5412 m/s / 0.015 s = 969.414 m/s^2

Answer 2

Sign convention: Down +ve.
Using v^2 = u^2 + 2 a s, v = ? with s = 4m, u = 0, and a = g = 10ms^-2,
velocity of hitting floor, v = squrt (2x10x4) = 8.9 ms^-1 = 8.9 ms^-1 downwards

A similar calculation for the upward motion
Up +ve
where v = 0, u = ? s = 1.65m , a = -10 ms^-2
gives the bouncing back velocity u = squrt( 2x 10x 1.65) = 5.7 ms^-1 UPWARDS

Now acceleration= Change in velocity / Time taken
Taking UP as +ve,

= (- 8.9 - 5.7)/15x10^-3
973 ms^-2

Roughly 100 times the g!!

OK?

Did you ever wonder what you would happen to you if you were inside such a HUGE bouncy ball ?

Answer 3

help me with sums... A ball is dropped from a top of 102 ft. each and every time it hits the floor, it rebounds to? Measles A ball is dropped from a top of 102 ft. each and every time it hits the floor, it rebounds to at least a million/2 its previous top. discover the finished distance it travels before coming to relax. This pastime me!! answer: top = H = 102 ft n ? H 0. ? H ..............................=H a million. ? H - H/2.......................=H/2 2. ? H/2 - H/2/2.................=H/4 3. ?H/4 - H/4/2. ................= H/8 4. ?H/8 - H/8/2 ..................=H/16 ? H/(2)^4 ? H/ (2)^n formula: enable n = style of leap Hf = very last leap H = Hf (a million -a million/2)^n 102 = a million (a million -a million/2)^n 102 = (a million/2)^n ==================== n = 6.673 ? style of leap till come to relax ==================== finished Distance Traveled = 102(a million +a million/2)^n finished Distance Traveled = 102(a million.5)^6.673 ====================================== finished Distance Traveled = 1526.36 ft ? Ans ====================================== wish this helps remember that Jesus loves you. comprehend Him in His words the Bible. God Bless Lim?E