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Leya [2.2K]
3 years ago
15

John bounces a ball off the ground, and it leaves the ground with a velocity of 10 m/s. How long will it take the ball to reach

maximum height?

Physics
2 answers:
Maksim231197 [3]3 years ago
8 0

Explanation:

This is true because at maximum height, the velocity is 0

wariber [46]3 years ago
3 0

Gravity makes an upward-moving object move 9.8 m/s slower every second.

So if it leaves the ground at 10 m/s, it'll run out of upward speed in

(10 m/s) / (9.8 m/s²)  =  <em>1.02 seconds </em>

At that time, since it has no more upward speed, it'll start falling. So at that instant, it was exactly at its maximum height.

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1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp
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Answer with explanation:

We are given that  

Mass of ball,m_1=75 g=\frac{75}{1000}=0075kg

1 kg=1000 g

Height,h_1=1.6 m

h_2=0.6 m

Horizontal velocity,v_x=2 m/s

Mass of platem_2=400 g=\frac{400}{1000}=0.4 kg

a.Initial velocity of plate,u_2=0

Velocity before impact=u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s

Where g=9.8 m/s^2

Velocity after impact,v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s

According to law of conservation of momentum  

m_1u_1+m_2u_1=-m_1v_1+m_2v_2

Substitute the values  

0.075\times 5.6+0=-0.075\times 3.4+0.4v_2

0.4v_2=0.075\times 5.6+0.075\times 3.4

v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s

Velocity of plate=1.69 m/s

b.Initial energy=\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J

Final energy=\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2

Final energy=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

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The intensity on a screen 20 ft from the light will be 0.125-foot candles.

<h3>What is the distance?</h3>

Distance is a numerical representation of the length between two objects or locations.

The intensity I of light varies inversely as the square of the distance D from the source;

I∝(1/D²)

The ratio of the intensity of the two cases;

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