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## Homework Statement

A runner jumps at 30 degrees to the ground and covers 8.90 m. What was the takeoff speed?

## The Attempt at a Solution

I have no idea where to start...

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- Thread starter SirMarksAlot
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- #1

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A runner jumps at 30 degrees to the ground and covers 8.90 m. What was the takeoff speed?

I have no idea where to start...

- #2

rl.bhat

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What happens to these components with respect to time?

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how would i go about solving the velocity for each component?

- #4

rl.bhat

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If v is the velocity what are the vertical and horizontal components?

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honestly, i have no idea

- #6

rl.bhat

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Open any text book. And refer the motion in two dimension.

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but all i have is the angle of the jump and the distance...no time or anything

- #8

rl.bhat

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Using proper equation you can find the remaining unknown quantities. Try to find out the equations.

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do u mean these formulas?

Fx = cos A * F1

Fy = sin A * F1

Fx = cos A * F1

Fy = sin A * F1

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honestly, im struggling and have no idea what to do...

- #11

rl.bhat

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Yes. These are the two components. Only replace F by v for the velocity.do u mean these formulas?

Fx = cos A * F1

Fy = sin A * F1

In that x component remains constant, because there is no acceleration in that direction. If t is the time of flight, then x = vcosθ*t.

In the vertical direction, the initial velocity is vsinθ. What is the final velocity when it reaches the maximum height. And what is the time taken to reach the maximum height?

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when u say x = vcosθ*t , what is v? the speed before the jump?

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rl.bhat

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- #15

rl.bhat

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Yes.

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I dont know how to find the time it would take to reach the max height.

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- #17

rl.bhat

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