<span><span>anonymous </span> 4 years ago</span>Any time you are mixing distance and acceleration a good equation to use is <span>ΔY=<span>V<span>iy</span></span>t+1/2a<span>t2</span></span> I would split this into two segments - the rise and the fall. For the fall, Vi = 0 since the player is at the peak of his arc and delta-Y is from 1.95 to 0.890.
For the upward part of the motion the initial velocity is unknown and the final velocity is zero, but motion is symetrical - it takes the same amount of time to go up as it does to go down. Physiscists often use the trick "I'm going to solve a different problem, that I know will give me the same answer as the one I was actually asked.) So for the first half you could also use Vi = 0 and a downward delta-Y to solve for the time.
Add the two times together for the total.
The alternative is to calculate the initial and final velocity so that you have more information to work with.
The source and the observer are moving towards each other. The observer is moving toward the source. The source is moving away from the observer
Answer:
a) 0.05s
b) 4000N
Explanation:
a)When car is stopped its final velocity become zero
U- 10 m/s
V- 0 m/s
S - 0.25 m
t -?
S = (v+u)*t/2
0.25 =(10+0)*t/2
t = 0.05s
b) If we happened to calculate the avarage force we have to consider about acceleration
V= 0
U = 10
t = 0.05 s
a =?
V = U + at
0 = 10 -a * 0.05
a = 200 m/s2
F = m *a
= 20 * 200
= 4000N
