A ball is thrown down vertically with an initial speed of 31 ft/s from a height of 40 ft. (a) What is its speed just before it s
1 answer:
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Answer:
a = - 1.987 × 10⁶ ft/s²
t = 6.84 × 10⁻⁴ s
Explanation:
v₀ = 910 ft/s
x = 5 in.
relation v = v₀ - k x
v = 0 as body comes to rest
0 = 900 - 5k/12
k = 2184 s⁻¹
acceleration
where
(A) a = -k × v
at v= 910 ft/s
a = - 1.987 × 10⁶ ft/s²
(B) at x = 3.9 in.
v = 910 - 3.9(2184)/12
v = 200.2 m/s
t = 6.84 × 10⁻⁴ s
Answer:
9.6 m/s
Explanation:
Angle of projection, θ = 28°
Horizontal distance, R = 7.8 m
Let the velocity of projection is given by u.
The formula used to find the velocity of projection is given by
u = 9.6 m/s
Thus, the velocity of projection is 9.6 m/s.
Answer:
<h2>9.39 m/s</h2>Explanation:
The velocity of the bowling ball can be found by using the formula
p is the momentum
m is the mass
From the question we have
We have the final answer as
<h3>9.39 m/s</h3>Hope this helps you