An athlete performing a long jump leaves the ground at a 27.0 degree angle and lands 7.80m away. what was the takeoff speed?

The initial velocity is 9.72 m/s.

Since an athlete performs a projectile motion;

R = u^2sin2θ/g

R = u^2sin2θ/g

Rg = u^2sin2θ

u^2 =Rg/sin2θ

u = √Rg/sin2θ

u = √7.80 × 9.8/sin(2)27.0

u = 9.72 m/s

The initial velocity is 9.72 m/s.

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=9.72 m/s

Explanation:

From the Newton's laws of motion;

x=2(v²cos∅sin∅)/g

Using geometry we see that 2 cos∅sin∅ = sin 2∅

Therefore, x= (v²sin 2∅)g,  where v is the take off speed x the range and ∅ the launch angle.

Making v the subject of the formula we obtain the following equation.

v=√{xg /(sin 2∅)}

x=7.80

∅=27.0

v=√{7.8×9.8/sin(27×2)}

v=√94.485

v=9.72 m/s

multiply 2kg by latent heat of vap