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liberstina [14]
3 years ago
6

If a ball is thrown vertically upward with a velocity of 80 ft/s, then its height after t seconds is s = 80t − 16t 2 . (a) what

is the maximum height reached by the ball? (b) what is the velocity of the ball when it is 96 ft above the ground on its way up? on its way down?
Physics
1 answer:
MAXImum [283]3 years ago
4 0
Max height occurs when v = 0.
v(t) = ds(t)/dt
v(t) = 80 - 32t
0 = 80 - 32t
t = 5/2

s(5/2) = 80(5/2) - 16(5/2)^2
s(5/2) = 100
Answer: 100 ft

96 = 80t - 16t²
t = 3, 2
(80 ± √256) / 32 using the quadratic equation.

v(2) = 16
v(3) = -16
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Answer:

F = 800N

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Given;

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The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed.

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F = (0.80×50)/0.05

F = 800N

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4 0
3 years ago
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vekshin1

kinematic equation

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Can anybody help me solve this problem? Thank you so much!
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