A ball is thrown vertically upward. What are its velocity and acceleration when it reaches its maximum altitude? What is its acc
1 answer:
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Answer:
(a) t = 1.14 s
(b) h = 0.82 m
(c) vf = 7.17 m/s
Explanation:
(b)
Considering the upward motion, we apply the third equation of motion:
where,
g = - 9.8 m/s² (-ve sign for upward motion)
h = max height reached = ?
vf = final speed = 0 m/s
vi = initial speed = 4 m/s
Therefore,
<u>h = 0.82 m</u>
Now, for the time in air during upward motion we use first equation of motion:
(c)
Now we will consider the downward motion and use the third equation of motion:
where,
h = total height = 0.82 m + 1.8 m = 2.62 m
vi = initial speed = 0 m/s
g = 9.8 m/s²
vf = final speed = ?
Therefore,
<u>vf = 7.17 m/s</u>
Now, for the time in air during downward motion we use the first equation of motion:
(a)
Total Time of Flight = t = t₁ + t₂
t = 0.41 s + 0.73 s
<u>t = 1.14 s</u>
Answer:
Explanation:
The spring is stretched by .5 m and then released that means its amplitude of oscillation A is 0.5 m .
A = 0.5 m
After the release at one extreme point , the mass comes to rest again at another extreme point after half the time period ie
T / 2 = .3 s
T = 0.6 s
Angular velocity
ω =
ω =
ω = 10.45
Maximum velocity = ω A
ω and A are angular velocity and amplitude of oscillation.
Maximum velocity = 10.45 x .5
= 5.23 m /s