<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.
Answer:
<h3>14.97m/s</h3>
Explanation:
Given
Initial velocity of the car u = 8m/s
Distance travelled by the rider S = 40m
Acceleration a = 2m/s²
Required
rider's velocity after the acceleration v
Using the equation of motion
v² = u²+2as
v² = 8²+2(2)(40)
v² = 64+160
v² = 224
v = √224
v = 14.97m/s
Hence the rider's velocity after the acceleration is 14.97m/s
Answer:
Explanation:
Given
See attachment for the graph
Required
Determine the frequency
Frequency (F) is calculated as:
Where
T = Time to complete a period
From the attachment, the wave complete a cycle or period in 3 seconds..
So:
--- Approximated
Answer:
distance = 3 + 4 = 7 mi
displacement is √(3² + 4²) = 5 mi
Explanation:
Answer:
Explanation:
To solve this problem we use the Hooke's Law:
(1)
F is the Force needed to expand or compress the spring by a distance Δx.
The spring stretches 0.2cm per Newton, in other words:
1N=k*0.2cm ⇒ k=1N/0.2cm=5N/cm
The force applied is due to the weight
We replace in (1):
We solve the equation for m:
