Finding initial velocity and theta, given range and height

In summary, the problem involves a stone being thrown down a slope with a range of 50m and a maximum height of 12m. The goal is to determine the magnitude of the initial velocity (V_o) and the initial angle (theta). To solve for V_o, the components of velocity (V_xo and V_yo) are broken down using the given equations, but the lack of information on time (t) presents a challenge. An alternative approach is to use the equation v_y^2=V_yo^2-g(y-yo) which does not involve time, but the initial angle (theta) has not been addressed yet.
  • #1
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Homework Statement



A stone is thrown down a slope. I am given the Range of the slope and the max height from spot thrown.
R= 50m
h= 12m

Homework Equations



I need to determine the magnitude of Vo and inital theta.

The Attempt at a Solution



First I need to find V_o:
I feel that the only relevant portion to find the initial velocity is from the spot thrown to the max height where V_yo is 0. I broke up the components of velocity so that:
V_xo=vcos(theta)
V_yo=vsin(theta)

However, I'm lost after this because I don't know t. My first idea was to find t, but I was having problems finding an equation that doesn't use v, theta, in a time equation.
My second idea was to find an equation that avoids time such as v_y^2=V_yo^2-g(y-yo).

I haven't attempted the initial theta part yet.
 
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  • #2
Reading through this post, this morning, I realized I forgot to add that the slope has a rise of 30m and a run of 40m. I don't feel that would help with find V_o though. :/
 
  • #3


Your approach is correct. To find the initial velocity, you can use the fact that at the maximum height, the vertical component of velocity is 0. This means that V_yo=0 and V_xo=Vcos(theta). Using the equation V_y^2=V_yo^2-2gh, you can solve for V. Then, using the equation V_x=Vcos(theta), you can solve for theta.
 

What is the equation for finding initial velocity and theta given range and height?

The equation for finding initial velocity and theta, given range and height, is:
v0 = √(g(R+h) / sin(2θ))
Where v0 is the initial velocity, g is the gravitational acceleration (9.8 m/s2), R is the horizontal range, h is the initial height, and θ is the angle of launch.

What are the required units for range, height, and initial velocity in this equation?

The units for range and height should be consistent (e.g. both in meters or both in feet). The unit for initial velocity will be m/s if the units for range and height are in meters, or ft/s if the units for range and height are in feet.

Can this equation be used for any object and any planet?

Yes, this equation can be used for any object and any planet, as long as the gravitational acceleration (g) is known. However, for objects on different planets, the value of g will be different and will need to be substituted in the equation accordingly.

How does the initial launch angle (θ) affect the initial velocity?

The initial launch angle (θ) affects the initial velocity by determining the vertical and horizontal components of the initial velocity. A higher launch angle will result in a higher vertical component of the initial velocity, while a lower launch angle will result in a lower vertical component. The horizontal component of the initial velocity will remain constant.

Can this equation be used for finding the initial velocity of objects launched at an angle from a height other than ground level?

Yes, this equation can be used for finding the initial velocity of objects launched at an angle from a height other than ground level. The initial height (h) should be added to the value of the range (R) in the equation.

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