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?

=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