Projectile Jumper Problem: Finding Takeoff Speed

[tuber313]

Homework Statement

A long jumper leaves the ground at 45 degrees above the horizontal and lands 8.2 meters away. What is her "takeoff" speed v0?

Homework Equations

x = x0 + v0t + 1/2at^2
v^2 = v0^2 + 2ad
v = v0 + at

The Attempt at a Solution

I can't seem to figure this problem out. Here's the information I got from the problem so far:

x component:

x0 = 0
x = 8.2 m
a = 0

y component:

y0 = 0
a = -9.8 m/s^2There seems to be too many missing variables to solve the equation. I even tried breaking the projectile in half, which yielded me this:

x component:

x0 = 0
x = 4.1 m
a = 0
v = v0

y0 = 0
a = -9.8 m/s^2
v = 0

EDIT: I forgot to mention I also realize that v0x and v0y components are equal, since the angle is 45 degrees. I know this is important, but I can't connect it to anything.

I still can't seem to figure it out, since I need y, or how high the object goes at its highest point. I also can't figure out the time it takes... If I knew either of those, I could figure it out, but I can't seem to get anything from this.

Thanks for your help.

 

[Clairefucious]

A few points you may find useful, for the full length problem

1) the jumper 'lands' - this means that at the end of the jump the y displacement is zero as she returns to the ground (y = 0)

2) the point above means that the final velocity in the y-direction must be equal in magnitude to the initial velocity but in the opposite direction (vy = -v0y)

3) you can solve for the time in the y-direction using v = v0 + at

I think you should be able to manage the rest. The trick is just to realize what's happened in the y-direction at the end of the jump.

 

[tuber313]

i understand that y=0 and vy=-v0y...

what i don't understand is how you can use v = v0 + at to find the time, since you still don't know what v0 (or v) is.

 

[Clairefucious]

A few points you may find useful, for the full length problem

1) the jumper 'lands' - this means that at the end of the jump the y displacement is zero as she returns to the ground (y = 0)

2) the point above means that the final velocity in the y-direction must be equal in magnitude to the initial velocity but in the opposite direction (vy = -v0y)

3) you can solve for the time in the y-direction using v = v0 + at

I think you should be able to manage the rest. The trick is just to realize what's happened in the y-direction at the end of the jump.

I should just mention that time will be a function of the v0y. When you use this time in the equation for the x-direction velocity you'll end up with something like v0x * v0y = number. Then just use what you know about the angles to get the answer.

 

[tuber313]

I solved it!

It wound up being 9.0 m/s, after I found the v0x=v0y=6.338769597

Thanks so much for your help!