To test the quality of a tennis ball, you drop it onto the floor from a height of 4m. It rebounds to a height of 2m. If the ball is in contact with the floor for 12ms, what is the magnitude of its average acceleration during that contact?
2007-09-18 09:57:05 · 1 answers · asked by sophia 2 in Science & Mathematics ➔ Physics
actually.. the answer is 1.26E3 m/s^2
:-/
2007-09-18 16:12:57 · update #1
the acceleration is related to F=m*a, where the net force is equal to the net m*a, so
First, calculate the speed of impact and the return speed using conservation of energy
since
.5*m*v^2=m*g*h
v=sqrt(2*g*h)
I will use g=10
On the descent,
vi=-sqrt(2*10*4)
vi=4*sqrt(5)
on the ascent
vo=sqrt(2*10*2)
vo=2*sqrt(10)
since you have the time of contact, then
2*sqrt(10)=4*sqrt(5)-a*12/1000
a=500*(2*sqrt(5)-sqrt(10))/3 m/s^2
a=220 m/s^2
j
2007-09-18 11:12:19 · answer #1 · answered by odu83 7 · 0⤊ 0⤋