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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?

Respuesta :

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

[tex]R=\frac{u^{2}Sin2\theta }{g}[/tex]

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

[tex]7.8=\frac{u^{2}\times 0.829}{9.8}[/tex]

u = 9.6 m/s

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

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