# A long jumper leaves the ground at an angle of 17 degrees to the horizontal and at a speed of 9.43 m/s. How far does he jump? What maximum height does he reach?Answer in units of m. The...

A long jumper leaves the ground at an angle of 17 degrees to the horizontal and at a speed of 9.43 m/s. How far does he jump? What maximum height does he reach?

Answer in units of m. The acceleration due to gravity is 9.8 m/s2.

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### 1 Answer

To solve this, you need to know how much of his velocity is horizontal and how much is vertical.

Construct a right triangle, starting at the origin, with an angle of 17 degrees, and a hypotenuse of 9.43 m/s. Then use sine and cosine to find the x- and y-values for velocity.

Sin 17 = y/9.43 m/s and y = 2.76 m/s vertical

cos 17 = x/9.43 m/s and x = 9.02 m/s

In the vertical direction the jumper has an initial velocity of 2.76 m/s but is slowing to a stop at the highest point because of the downward acceleration of gravity.

Since Vf = Vi + gt, where Vf is the final velocity which is zero; Vi is the initial velocity which is 2.76 m/s and g is a -9.8 m/s/s if up is defined as the positive direction.

0 = 2.76 m/s + (-9.8 m/s/s) t

t = .282 seconds up + .282 seconds down for a total time in the air of 0.563 s,

The entire time the jumper is in the air he is moving horizontally at 9.02 m/s. Since d = vt, d = 9.02 * 0.563 = 5.08 m jump.

The maximum height in the y-direction is given by:

delta y = Vit + 1/2 gt^2

delta y = 2.76 m/s * 0.282 s + 1/2 (-9.8 m/s/s)(0.282^2)

delta y = 0.39 m high

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