Question

While competing in the long jump, a person leaps over a smooth horizontal sand surface. She lands on the surface with speed vf = 8.25 m/s at an angle θ = 43.5° below horizontal. Assume that the person moves without air resistance. Use a Cartesian coordinate system with the origin at her final position. The positive x-axis is directed from her initial to her final position, and the positive y-axis is directed vertically upwards. What x position, in meters, did the jumper begin her long jump?

Answer 1

Answer:

here the horizontal displacement of the person while he jump is 6.93 meter.

Explanation:

As we know that the motion of the person is same as projectile motion from ground to ground

So here we can say that initial and final velocity is same in magnitude and also the angle will be same with the horizontal

So here the displacement in x direction is horizontal range that it will cover

So we will have

$R = \frac{v^2sin2\theta}{g}$

$R = \frac{8.25^2 sin2(43.5)}{9.81}$

$R = 6.93 m$

So here the horizontal displacement of the person while he jump is 6.93 meter.