Projectile motion given theta and x

[tkahn6]

Homework Statement

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

Homework Equations

8.7m = cos(23^o)V_ot

y = sin(23^o)t + (1/2)(-9.8_m/s²)t²

0_m²/s² = sin²(23^o)V_o² + 2(-9.8_m/s²)y

The Attempt at a Solution

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

The answer in the book is 11_m/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!

 

[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

 

[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...

 

[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