Question

You are a member of a citizen's committee investigating safety in the high school sports program. You are interested in knee damage to athletes participating in the long jump (sometimes called the broad jump). The coach has her best long jumper demonstrate the event for you. He runs down the track and, at the take-off point, jumps into the air at an angle of 30 degrees from the horizontal. He comes down in a sand pit at the same level as the track 26 feet away from his take-off point. With what velocity (both magnitude and direction) did he hit the ground?

Answer 1

Answer:

The long jumper hits the ground with a velocity of 158.54 ft/s at and angle of 30° from the horizontal.

Explanation:

This is a an example of projectile motion.

It should first be stated that the effects of air resistance and friction are negligible, so horizontal velocity does not change over the course of the jump.

To solve this problem, let's first express the x and y velocity of the jumper as components of the initial velocity and angle:

$V_x$ = V*Cos(30°)

$V_y$ = V*Sin(30°)

The distance traveled by the athlete is 26 feet, so we have:

Distance = Speed * time

26 = V*Cos(30) * time         -Equation (1)

We can also make an equation for the y velocity as follows:

$s=V_y*t + 0.5(a*t^2)$

here a = -32.2 ft/s^2

s = 0 ft

and $V_y$=V*Sin(30°)

so we get:

$0=VSin(30)*t+0.5(-32.2t^2)$            -Equation (2)

Solving equations 1 and 2 simultaneously we get:

V = 158.54 ft/s

t = 0.1894 s

The direction of landing remains the same as the direction of the initial jump (30° from the horizontal). This is because the height is the same, and horizontal velocity remains constant with vertical velocity being exactly the same negative value as that from liftoff.