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A football is kicked straight up into the air; it hits the ground 5.2s later.

A. What was the greatest height reached by the ball? Assume it is kicked from ground level.

B. With what speed did it leave the kicker's foot?

How do i solve these? i dont want just an answer i want the process so i can memorize it

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A football is kicked straight up into the air; it hits the ground 5.2s later.

When an object moves upward, its velocity decreases 9.8 m/s each second, until the velocity is 0 m/s.

When the velocity = 0 m/s, the object is at its highest position.

Then the object falls back to its initial position. As it falls, its velocity increases 9.8 m/s each second.

I always use the down trip, because the initial velocity – 0 m/s!!!

Initial velocity up = Final velocity down

The distance up = distance down

The time up = time down

Total time = time up + time down = 5.2 s

Time down = ½ * total time

Time down = ½ * 5.2 s = 2.6 s

Time down =2.6 s

Distance = Initial velocity * time + (½ * acceleration * time ^2)

Initial velocity for down trip = 0 m/s

Acceleration = 9.8 m/s

Time = 2.6 s

Distance = 0 * time + (½ * 9.8 * 2.6 ^2) = 33.124 m

Final velocity down = Initial velocity + (acceleration * time)

Final velocity down = 0 + (9.8 * 2.6) = 25.48 m/s

Initial velocity up = 25.48 m/s

A. What was the greatest height reached by the ball? Assume it is kicked from ground level. 33.124 m

B. With what speed did it leave the kicker's foot? 25.48 m/s
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A. In theoretical physics, it is theorized that it takes exactly the same amount of time to go down as it went up. So we can divide by 2 to figure out the time it took for the football to reach maximum height.

T = 5.2 / 2 = 2.6 s

Vf = Vi + AT

Vi = Vf - AT

Vi = (0) - (-9.8)(2.6) (Vf is zero because the ball has zero velocity at max height)

Vi = 25.48

Vf^2 - Vi^2 = 2ΔYA

ΔY = (Vf^2 - Vi^2) / 2A

ΔY = ((0)^2 - (25.48)^2) / ((2)(-9.8))

ΔY = 33.12 m

B. We already solved for the balls initial velocity in order to reach its maximum height. The answer was 25.4 m/s

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A

5.2/2=2.6 this is the time it take the foot ball to reach maximum height.

Calculate the distance it falls from it's maximum height. First write down all the values you know

initial velocity(Vi)=0 (no vertical motion at the top of it's maximum height, it "stops before it starts to fall downward.

a=9.8, this is due to gravity. t= 2.6 seconds. We want to find d, distance. Use a kinematic equation will these factors in it. In this case. d=Vit+0.5at^2

because Vi=0 Vit=0 therefor in this case d=0.5at^2

Substitute in the values. d=0.5x9.8x2.6^2=33.124m

You'll probably want to round this to 33m

B

You need to use the kinematic equations again for this one. Write down the factors you know, and the ones you want to find out.

For the motion, from being hit by the foot, to reaching maximum point.

a= -9.8 (gravity causing ball to decelerate that's why it's negative)

Vf=0

d=33m

t=2.6 seconds

Vi=? this is what we want to find.

this formula will word. Vf^2=Vi^2 +2ad

Vf=0 so Vf^2=0

rearrange to find Vi

Vi=Square rt of -2ad

Vi=Square rt of -2x-9.8x33

Vi=Square rt of 646.8

Vi=25.4 m/s

Good luck!

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