Projectile Motion High Jumper

In summary: Using the range equation d = vxt, we can solve for vxi. Then using the time equation t = \frac {vy}{a} and plugging in the values for d and a, we can solve for vyi. The initial velocity of the high jumper must be 6.99 m/s at an angle of 63 degrees to clear the bar. Assumptions made include constant acceleration due to gravity, no air resistance, and a perfectly executed jump. In summary, to determine the initial velocity required for a high jumper to clear a 2.0 m bar with a range of 2.0 m, we can use the equations for projectile motion and make assumptions of
  • #1
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Physics Kinematics Unit: Question concering Projectile Motion

Homework Statement


If a high jumper reaches her maximum height as she travels across the bar, determine the initial velocity she must have to clear a bar set at 2.0 m if her range during the jump is 2.0 m. What assumptions did you make to complete the calculations?

Range(d) = 2.0 m
Acceleration due to gravity (a) = 9.81 m/s^2 down
The height of the bar = 2.0 m

Homework Equations



[tex] a= \frac{Vf - Vi}{t} [/tex]

[tex] d=(Vi)(t) + (1/2)(a)(t^2) [/tex]

[tex] d=Vt [/tex]



The Attempt at a Solution


This is my attempt. Since Vix is O, we can determine time using this equation:
[tex] d=(Vi)(t) + (1/2)(a)(t^2) [/tex]

[tex] 2.0 m = (0)(t) + (1/2) (9.81 m/s^2) (t^2)[/tex]

[tex] t = 0.64 s[/tex]

Now trying to find Vix, use this equation:

[tex] d = vt[/tex]

[tex] 2.0 m = V(0.64) [/tex]

[tex] v = 3.13 \frac {m}{s} [/tex]

Now You have to divide the time by 2, because that's half way through the jump, her maximum height would be half the time.

[tex] t = \frac {0.64}{2} [/tex]

[tex] t = 0.32 s [/tex]

Now find Viy, using this equation:

[tex] d = Vt[/tex]

[tex] 2.0 = V(0.32)[/tex]

[tex] v = 6.2 \frac {m}{s}[/tex]

Now find Vi using this this equation:

[tex] a^2 + b^2 = c^2 [/tex]
[tex] 6.25^2 + 3.13 ^2 = c^2[/tex]
[tex]c= 6.99 m/s (63 degrees)[/tex]

I got this wrong, but I'm just showing you my attempt to this question
 
Last edited:
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  • #2
Motion of the high jumper is the projectile motion.
So vyi = vsin(theta) and vxi = vcos(theta)
 
  • #3
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Your attempt at the solution is a good start, but there are a few things that need to be clarified. Firstly, the equation d = vt only applies to objects moving with constant velocity, not to projectiles like the high jumper. In this case, the jumper is accelerating due to gravity, so we cannot use this equation.

Secondly, the equation a = (Vf - Vi)/t is not applicable in this situation either. This equation is used for calculating acceleration, but the high jumper's acceleration is constant (due to gravity), so we do not need to use this equation.

To solve this problem, we can use the equations you started with, but we need to use them correctly. The equation d = (Vi)(t) + (1/2)(a)(t^2) is the correct equation to use, but we need to solve for Vi, not t. This can be done by rearranging the equation to be Vi = (d - (1/2)(a)(t^2))/t. Plugging in the given values, we get Vi = (2.0 m - (1/2)(9.81 m/s^2)(0.64 s)^2)/(0.64 s) = 3.13 m/s. This is the initial velocity in the x-direction, which is what we were asked to find.

For the initial velocity in the y-direction, we can use the equation Vf = Vi + at. Since the final velocity in this case is 0 (at the maximum height), we can rearrange the equation to be Vi = -at. Plugging in the given values, we get Vi = -(9.81 m/s^2)(0.32 s) = -3.13 m/s. This means that the initial velocity in the y-direction is downwards, which makes sense since the jumper is accelerating downwards due to gravity.

To find the total initial velocity, we can use the Pythagorean theorem as you did. However, we need to use the correct values. The side lengths of the triangle are 3.13 m/s (x-direction) and -3.13 m/s (y-direction). Plugging these into the equation a^2 + b^2 = c^2, we get c = 4.43 m/s. This is the total initial velocity of the jumper.

As for assumptions, the calculations made here assume that the jumper is
 

1. What is projectile motion in the context of high jumping?

Projectile motion refers to the motion of an object that is thrown or launched into the air and is affected by both horizontal and vertical forces, such as gravity. In high jumping, the athlete's body can be considered a projectile as they jump over the bar.

2. How does the high jumper's body move during a jump?

During a high jump, the athlete's body follows a parabolic trajectory, meaning it follows a curved path. The athlete starts with an upward vertical velocity, reaches the highest point of the jump, and then falls back to the ground.

3. What factors affect the trajectory of a high jumper?

The trajectory of a high jumper is affected by several factors, including the takeoff angle, the athlete's initial velocity, air resistance, and gravity. These factors can be manipulated by the athlete to achieve a higher jump.

4. How does air resistance impact a high jumper's performance?

Air resistance, also known as drag, can slow down the motion of a high jumper. The higher the athlete jumps, the more air resistance they will face, making it more challenging to clear the bar. Athletes can minimize the impact of air resistance by using techniques such as the Fosbury Flop.

5. What is the optimum takeoff angle for a high jumper?

The optimum takeoff angle for a high jumper is typically between 20 and 30 degrees, depending on the athlete's height and technique. This angle allows the athlete to achieve the longest horizontal distance while still gaining enough height to clear the bar.

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