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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?

Answer in units of m/s.

Added Mon, 15 Jun '15

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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.

Tangent = opposite / adjacent.

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.

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

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

t = 16.099 m / (16 m/s * cos(57)) = 1.847 s

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

Added Mon, 15 Jun '15

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