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Fynjy0 [20]
2 years ago
15

a ball is thrown straight up with an initial velocity of 20.0 ft/s. how long dose it take to reach maximum height?​

Physics
1 answer:
timama [110]2 years ago
5 0

Answer:

At the maximum height the ball will have an instantaneous velocity of zero. The formula for instantaneous velocity is

v = v₀ + at

Where v is the instantaneous velocity, v₀ is the initial velocity, a is the acceleration due to gravity, -9.8 m/sec², and t is the time.

We need instantaneous velocity to = 0.

0 = 20 - 9.8t : Isolate t => 1st step: Add 9.8t to both sides

9.8t = 20 : Isolate t => 2nd step: Divide both sides by 9.8

t = 2.04 seconds (approximately)

Explanation:

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Answer:

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Explanation:

7 0
3 years ago
I need help with this. I think I got it correct, but I wanna make sure.
kicyunya [14]

Answer:

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Explanation:

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3 years ago
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A 86g ball is dropped vertically to the floor from a height of 2.87m and bounces to a height of 1.28. What is the magnitude of t
irga5000 [103]

Answer:

The impulse received by the ball from the floor during the bounce is approximately 1.11329438 m·kg/s

Explanation:

The given mass of the ball, m = 86 g = 0.089 kg

The height from which the ball is dropped, H = 2.87 m

The height to which the ball bounces, h = 1.28 m

Mathematically, we have;

Δp = F·Δt

Where;

Δp = The change in momentum = m·Δv

F = The applied force

Δt = The time of contact with the force

The velocity of the ball just before it touches the ground, v₁ = -√(2·g·H)

The velocity with which the ball leaves, v₂ = √(2·g·h)

The change in momentum, Δp = m·(v₂ - v₁)

∴ Δp = m·(√(2·g·h) - (-√(2·g·H))) = m·(√(2·g·h) +√(2·g·H) )

The impulse, Δp, received by the ball from the floor during the bounce is given as follows;

Δp = 0.089 kg × (√(2 × 9.8 m/s² × 1.28 m) + √(2 × 9.8 m/s² × 2.87 m)) ≈ 1.11329438 m·kg/s

The impulse received by the ball from the floor during the bounce, Δp ≈ 1.11329438 m·kg/s

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Answer:

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Contact forces is the answer

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