Kinematics: Projectile with no initial velocity

[huynhtn2]

Homework Statement

A person wants to shoot a rocket off of a mountain 3.70 km high so that the rocket, when shot at 37.7° above the horizontal from the very top of the mountain, lands at the foot of the mountain 9.65 km (horizontal distance) away. What initial speed must the rocket have? (Assume that the only force acting on the rocket after it is blasted from the top of the mountain is gravity.)

Homework Equations

d=vit + 0.5at^2
vf=vi +at
vf^2=Vi^2 +ad

The Attempt at a Solution

I drew out the diagram and labeled all the angles on the mountain ( i don't know if i need them). I really don't know where to go from here. Can someone help me out? I tried making equations but there are always two unknown values.

 

[Delphi51]

Call the initial speed v. Then use a bit of trigonometry to get the vertical and horizontal parts of v. From here on, you do the horizontal and vertical motions separately because one does not affect the other. The only thing they have in common is the time of flight. In the horizontal part, you have constant speed so use d = vt. Use your accelerated motion formulas for the horizontal part. Put in your numbers and you'll find you can solve two of the 3 formulas as a system of equations to find the v.

 

[huynhtn2]

so
vcostheta = horizontal velocity
vsintheta = vertical velocity
then
d = vcostheta(t)

how do i get the equation for the vertical component? doe i do
vf = vi +at
0 = vsintheta + (-9.81)t?

 

[huynhtn2]

Call the initial speed v. Then use a bit of trigonometry to get the vertical and horizontal parts of v. From here on, you do the horizontal and vertical motions separately because one does not affect the other. The only thing they have in common is the time of flight. In the horizontal part, you have constant speed so use d = vt. Use your accelerated motion formulas for the horizontal part. Put in your numbers and you'll find you can solve two of the 3 formulas as a system of equations to find the v.

so
vcostheta = horizontal velocity
vsintheta = vertical velocity
then
d = vcostheta(t)

how do i get the equation for the vertical component? doe i do
vf = vi +at
0 = vsintheta + (-9.81)t?

 

[Delphi51]

d = v*cos(θ) looks good!
The vf at the bottom of the mountain will not be zero!
Better switch to the distance formula for the vertical motion because you know the vertical distance.

 

[huynhtn2]

d = v*cos(θ) looks good!
The vf at the bottom of the mountain will not be zero!
Better switch to the distance formula for the vertical motion because you know the vertical distance.

i see, but that distance doesn't include the vertical parabola time that the projectile goes through. I don't know how to include that part.

 

[Delphi51]

In d = Vi*t + .5*a*t², the t is the time for the change in distance d.
You must replace d with -3700 m. The curve of the parabola has nothing to do with it because the vertical and horizontal motions are completely independent. The formula takes care of the up and down business - don't worry about it!

 

[huynhtn2]

In d = Vi*t + .5*a*t², the t is the time for the change in distance d.
You must replace d with -3700 m. The curve of the parabola has nothing to do with it because the vertical and horizontal motions are completely independent. The formula takes care of the up and down business - don't worry about it!

I see, Thanks your your help Delphi! :)

 

[Delphi51]

Most welcome! I enjoyed the problem.