Question

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?

Answer 1

To find the takeoff speed of an athlete performing a long jump, we can use the principles of projectile motion.

Step-by-step explanation:

To calculate the takeoff speed of an athlete performing a long jump, we can use the principles of projectile motion. Since the jump is treated as a projectile, we can break down the motion into horizontal and vertical components.

First, we need to find the vertical component of the takeoff speed. We know that the jumper leaves the ground at a 27.0 degree angle and lands 7.80m away. The vertical displacement can be calculated using the equation:

Δy = v0y * t - (1/2) * g * t^2

The horizontal component of the takeoff speed can be found using the equation:

Δx = v0x * t

By combining these two equations and solving for the takeoff speed, we can determine the answer.

Answer 2

=9.72 m/s

Step-by-step 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