# Projectile Motion High Jumper

1. Apr 8, 2009

### Exquisite_

Physics Kinematics Unit: Question concering Projectile Motion

1. The problem statement, all variables and given/known data
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

2. Relevant equations

$$a= \frac{Vf - Vi}{t}$$

$$d=(Vi)(t) + (1/2)(a)(t^2)$$

$$d=Vt$$

3. The attempt at a solution
This is my attempt. Since Vix is O, we can determine time using this equation:
$$d=(Vi)(t) + (1/2)(a)(t^2)$$

$$2.0 m = (0)(t) + (1/2) (9.81 m/s^2) (t^2)$$

$$t = 0.64 s$$

Now trying to find Vix, use this equation:

$$d = vt$$

$$2.0 m = V(0.64)$$

$$v = 3.13 \frac {m}{s}$$

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.

$$t = \frac {0.64}{2}$$

$$t = 0.32 s$$

Now find Viy, using this equation:

$$d = Vt$$

$$2.0 = V(0.32)$$

$$v = 6.2 \frac {m}{s}$$

Now find Vi using this this equation:

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

I got this wrong, but I'm just showing you my attempt to this question

Last edited: Apr 8, 2009
2. Apr 8, 2009

### rl.bhat

Motion of the high jumper is the projectile motion.
So vyi = vsin(theta) and vxi = vcos(theta)