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

The long jumper's starting x position is 6.97 meters from her landing point, calculated using projectile motion principles and components of final velocity.

Step-by-step explanation:

To calculate the x position from where the long jumper began her jump, we need to apply the principles of projectile motion. Because there is no air resistance, we can treat the horizontal and vertical motions separately.

The horizontal displacement can be found using the formula x = vxt, where vx is the horizontal component of the final velocity and t is the time of flight. However, we're not given the time directly, so we have to calculate it using the vertical component of the motion.

First, we find the horizontal component vx and the vertical component vy of the final velocity:

vx = vfcos(θ) = 8.25 m/s * cos(43.5°) = 8.25 m/s * 0.731 = 6.03 m/s

vy = vfsin(θ) = 8.25 m/s * sin(43.5°) = 8.25 m/s * 0.687 = 5.67 m/s

The time t in the air is the same for both the ascent and descent of the jump. Since the landing point is at the same vertical level as the take-off point, the time to reach the maximum height is half of the total flight time. The vertical motion can be described by vy = gt, where g is the acceleration due to gravity (9.81 m/s2). Therefore, we get t = vy/g = 5.67 m/s / 9.81 m/s2 = 0.578 s. This is the time to reach maximum height, so the total time is double this: ttotal = 0.578 s * 2 = 1.156 s.

Having the total time of flight, we can now compute the horizontal displacement:

x = vxttotal = 6.03 m/s * 1.156 s = 6.97 m

Therefore, the x position where the jumper started is 6.97 meters from the point of landing.

Answer 2

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

Step-by-step 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 = (v^2sin2\theta)/(g)`

`R = (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.