Question

A jumper in the long-jump goes into the jump with a speed of 12 m/s at an angle of 37 degrees, what is the range?

Answer 1

So we need two steps to solve this. First, we need to use the vertical information to see how long the jumper is in the air. Then, we can use that time frame to gauge how far he went due to his speed at the jump.

Solving for time:
Y = Yo + Voy * t - 0.5 * g * t^2
Y = 0 m
Yo = 0 m
Voy = 12sin(37) = 7.22 m/s
g = 9.8 m/s^2
t = ?

0 = 0 + 7.22t - 4.9t^2
-4.9t^2 + 7.22t = 0
t (-4.9t + 7.22) = 0
t1: 0 s
t2: -4.9t + 7.22 = 0
-4.9t = -7.22
t2 = 1.47 s

Obviously the jumper isn't flying for 0 s, so we reject that and take t2 = t = 1.47 s

Now we know once the jumper is flying, he is not acted on in the horizontal direction. So his speed will be constant. If we take the product of speed and time, we can find the range of the jump.

Solving for the range:
x = v*t
t = 1.47 s
v = 12cos(37) = 9.58 m/s

x = (9.58)(1.47) = 14.1 m

So the range of the jump is 14.1 m