Finding initial speed given two angles and a distance

[heyhowsitgoin]

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 doesn't work

 

[rl.bhat]

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.

 

[heyhowsitgoin]

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

 

[rl.bhat]

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

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

 

[heyhowsitgoin]

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]

It is right. Now simplify everything. And solve for Vo.

 

[LiquidSword]

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]

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