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Jimmy is at the bottom of a hill while Billy is on the hill a distance 18 m up the hill from the bottom.
Jimmy is at the origin of an xy coordinate system, and the line that follows the slope of the hill has an equation y = (0.5) x.
Jimmy throws an apple to Billy at an angle of 57 degrees with respect to the horizontal.
The acceleration of gravity is 9.8 m/s^2.

With what speed v_0 must he throw the apple to reach Billy?

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Givens:
x = 18m = distance up hill
(y = 0.5x) = slope of hill
57 degrees  = angle of throw

Find the angle of the hill.
We can do this by using trigonometry.
y = 0.5x tells us that when y = 2, x = 1.
In other words when taking the angle at the base of the hill, the opposite side = y = 1, and the adjacent side = x = 2.

Which means :
Tan ( theta ) = 1 / 2
theta = arctan(1/2)
theta = 26.57 degrees.

Use Kinematic equations to describe the system.
S_x = distance in X direction till Billy.
S_x = 18m * cos(26.57) = 16.099 m
S_x = V_x * t = V * cos(57) * t   ------ (kin. equation)

S_y = distance in Y direction till Billy
S_y = 18m * sin(26.57) = 8.051 m
S_y = V_y *t + 1/2 * (-g) * t^2
= V * sin(57) * t + 1/2 (-g) * t^2 ----- (kin. equation)

We have Two Unknowns ( t and V ) and two equations. Because the equations are part of the same system, we can solve them.

Solve for V
S_x = V * cos(57) * t
t = S_x / ( V * cos(57) )

S_y = V * sin(57) * t + 1/2 * (-g) * t^2

(plut in t from the equation above)
S_y = V * sin(57) * (S_x / (V*cos(57) ) + 1/2 * (-g) * (S_x / (V*cos(57) ) ^2
= V / V * sin(57) * S_x/cos(57) + 1 / V^2 * 1/2 * (-g) * (S_x)^2 / cos(57)^2

S_y - S_x * sin(57) / cos(57) = 1/2 * (-g) * (S_x)^2 / V^2 / cos(57)^2

V^2 = (-g) * (S_x)^2 / ( 2 * cos(57)^2 * (S_y - S_x * sin(57) / cos(57) )
V^2 = -9.81 m/s^2 * (16.099 m)^2 / ( 2 * cos(57)^2 * (8.051 m - 16.099 m * sin(57) / cos(57) )
V^2 = -2542.53 m^3/s^2 / ( 0.593 * (-16.739m) )
V^2 = 256.14 m^2 / s^2
V = 16.00 m/s

Double Check our Work.
t = S_x / ( V * cos(57) )
t = 16.099 m / (16 m/s * cos(57)) = 1.847 s

does y(1.847) = S_y ?

y(t) = V * sin(57)*t + 1/2 (-g) * t^2

y(1.847) = 16.00 m/s * sin(57) * (1.847 s) + 1/2 * -9.81 m/s^2 * (1.847 s)^2
= 24.78 m - 16.73 m = 8.047

8.047 ~= 8.051
Checks out.
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