02 Finals Tennis
Physics hw question help 3(final)?
A tennis ball is dropped from 1.11 m above the ground. It rebounds to a height of 1.02 m. (Assume that the positive direction is upward.)
(a) With what velocity does it hit the ground?
(b) With what velocity does it leave the ground?
(c) If the tennis ball were in contact with the ground for 0.010 s, find its acceleration while touching the ground.
(d) Compare the acceleration to g.
This is the last one i need help with for this assignment
i under stand how u did part a and b and d but i don’t get what u do for c
Hello
a) use s = 1/2gt^2 and v= g*t to get s = 1/2 v^2/g –> v^2 = 2sg –>
v = √2sg = √(2*1,11m*9,81 m/sec^2)
v = 4,55 m/sec (hitting the ground)
b) the height reached is h = v^2/2g
v^2 = h*2g = 1,02m *2*9,81 m/sec^2
v = 4,47 m/sec (leaving the ground)
c) the acceleration was 4,55 m/sec/0,01 sec = (-) 455 m/sec^2 when coming down and being stopped,
and 4,47m/sec/0,01 sec = 447 m/sec^2 when accelerated up (if both took 0,01 sec). Compare to 9,81 m/sec^2 of the gravitational acceleration.
Regards
Hi again
Acceleration is increase of velocity per second.
Wen the ball is down on the ground the velocity is for a very short moment at zero .
Then the velocity increases to 4,5 m/sec within 0,01 second. So the acceleration is (4,5 m/sec)/0,01 second.
All right? But acceleration is not defined as increase of velocity within 0,01 second, but within 1 second.
So with the same acceleration of 4,5 m/sec in 0,01 seconds, the increase of velocity within 1 second would be 100 times as large = (450 m/sec)/sec, which is 450 m/sec^2.
Therefore, you calculate the acceleration as velocity divided by the time during which this velocity was reached, i.e., (4,5 m/sec)/ 0,01 second
= 450 m/sec^2.
And when the ball comes from above and stops on the ground within 0,01 second, the acceleration is identical, except that it is negative acceleration.
If it was meant that the ball is a total of 0,01 seconds in contact with the ball while decreasing its velocity from 4,5 m/sec to zero and increasing it again to about 4,5 m/sec, the the acceleration would be even twice as high.
