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Lesechka [4]
3 years ago
10

An athlete executing a long jump leaves the ground at a 28.0º angle and travels 7.80 m. (A) What was the takeoff speed?

Physics
1 answer:
Naddika [18.5K]3 years ago
7 0

Answer:

9.6 m/s

Explanation:

Angle of projection, θ = 28°

Horizontal distance, R = 7.8 m

Let the velocity of projection is given by u.

The formula used to find the velocity of projection is given by

R=\frac{u^{2}Sin2\theta }{g}

7.8=\frac{u^{2}Sin\left (2\times 28  \right ) }{9.8}

7.8=\frac{u^{2}\times 0.829}{9.8}

u = 9.6 m/s

Thus, the velocity of projection is 9.6 m/s.

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Let's first calculate the maximum height reached by using third equation of motion.

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The time it will take for the rocket to return to ground level will be 2t.

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