Answer:
(a) 5.714s
(b) 0.625m
Step-by-step explanation:
Acceleration due to gravity=a=±9.8m/s/s (+ when falling, - when going upwards)
(a)
STAGE A: between start and first bounce
Initial speed=u=0m/s, Distance=s=40m, and Final velocity=v
v²=u²+2as
v²=0²+2(10)(40)
v²=794
v=28m/s (As the ball hits the ground)
v=u+at
28=0+9.8t[tex]_{a}[/tex]
t[tex]_{a}[/tex]=28/9.8=2.85714285714s
STAGE B: from the first bounce until it starts to fall
u=half the speed it hit the ground with=(28/2)=14m/s, and v=0m/s(when it starts to fall)
v=u+at
0=14-9.8t[tex]_{b}[/tex]
t[tex]_{b}[/tex]=14/9.8=1.42857142857s
STAGE C: between when it starts to fall after bounce 1, and bounce 2
We can assume that the time it takes to go from the ground to max height after bounce 1 is equal to the time it takes to fall from that same height to the ground. Therefore, t [tex]_{c}[/tex]=1.42857142857s
The time from when the ball was released to when it hit the ground for the second time = t
t = t[tex]_{a}[/tex] + t[tex]_{b}[/tex] + t[tex]_{c}[/tex]
t=2.85714285714+1.42857142857+1.42857142857
t=5.71428571428s≈5.714s
(b)
Before bounce 1, u=28m/s (see stage a)
After bounce 1, u= 28/2= 14m/s
After bounce 2, u= 14/2 = 7m/s
After bounce 3, u= 7/2 =3.5m/s
u=3.5m/s and v=0m/s(when it starts to fall)
v²=u²+2as
0²=3.5²+2(-9.8)s
19.6s=12.25
s=0.625m (max height after third bounce)
