# Homework Help: Projectile motion given theta and x

1. Jul 5, 2009

### tkahn6

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

An Olympic long jumper leaves the ground at an angle of 23o and travels through the air for a horizontal distance of 8.7m before landing. What is the takeoff speed of the jumper?

2. Relevant equations

8.7m = cos(23o)Vot

y = sin(23o)t + (1/2)(-9.8m/s2)t2

0m2/s2 = sin2(23o)Vo2 + 2(-9.8m/s2)y

3. The attempt at a solution

Fifteen minutes and many permutations later, I get t = .835, .782 with Vo = 11.32m/s, 12.09m/s

The answer in the book is 11m/s.

Can you explain the steps you would take to solve this? It literally took me 15 minutes of mathematical manipulation to isolate t. The final step to find t for me was:

.697 = t2(3.6934 - 4.9t2)

Thanks guys!

2. Jul 5, 2009

### Dweirdo

y = sin(23)*V0*t + (1/2)(-9.8m/s2)t
u missed a Vo here.
well get t from the first equation, and substitute it in t in the second one.
from the first equation u get
t=8.7 / cos(23)Vo

3. Jul 5, 2009

### ideasrule

An easier way is to solve for t first in y = sin(23)*V0*t + (1/2)(-9.8m/s2)t^2:

0=sin(23)*V0*t + (1/2)(-9.8m/s2)t^2
Divide out t...

4. Jul 6, 2009

### andrevdh

$$x=V_o \cos(\theta_o)t$$

giving

$$t = x\frac{x}{V_o \cos(\theta_o)}$$

and from

$$y = V_o \sin(\theta_o)t - 0.5gt^2$$

we have that

$$y = x\tan(\theta_o) - \frac{gx^2}{2V_o\cos^2(\theta_o)}$$

the parabolic equation describing the trajectory of the projectile

Last edited: Jul 6, 2009