# Finding initial speed given two angles and a distance

## Homework Statement

A child runs down a 11 degree hill and then suddenly jumps upward at a 16degree angle above the horizontal and lands 1.2 m down the hill as measured along the hill

What was the child's initial speed?

## Homework Equations

Projectile motion

x = Vx*t

y = yo + Vyo*t - 1/2gt^2

## The Attempt at a Solution

tried solving for time t from the vertical motion equation y = yo+ Vyo*t - 1/2gt^2 and plug it in the
x = vx*t..but it doesnt work

Last edited:

rl.bhat
Homework Helper
Hi heyhowsitgoin
Welcome to PF.
If you draw the figure of inclined plane and the landing of the child, you can see that
y = 1.2*sin11.
t = x/vx = 1.2*cos11/vo*cos16
vyo = vo*sin16.
Substitute these values in the equation and solve for vo.

Hi thanks for the reply. Which equation are you referring to solve for vo?

rl.bhat
Homework Helper
Hi thanks for the reply. Which equation are you referring to solve for vo?

y = yo+ Vyo*t - 1/2gt^2

hrmm ok

so i end up with 0.229 = 0 + Vo*sin16 (1.2*cos11 / Vo*cos16) - 1/2(9.8)(1.2*cos11/Vo*cos16)^2

is this right? not sure how to solve for Vo at this point.

rl.bhat
Homework Helper
It is right. Now simplify everything. And solve for Vo.

Hello,

I was looking at this question and was not able to isolate Vo in the final equation. I ended up with a negative number under a square root...
Any suggestions on how to correctly isolate Vo?

rl.bhat
Homework Helper
hrmm ok

so i end up with 0.229 = 0 + Vo*sin16 (1.2*cos11 / Vo*cos16) - 1/2(9.8)(1.2*cos11/Vo*cos16)^2

is this right? not sure how to solve for Vo at this point.
In the equation final position y is zero, and initial position yo is 0.229