How Does Projectile Motion Determine the Initial Velocity for a High Jumper?

[Exquisite_]

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

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

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

d=Vt

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

 

[rl.bhat]

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