A ball is launched with a velocity of 10 m/s at an angle of 50.0 degrees to the horizontal. The launch point is at the base of a ramp of horizontal length 6.00 m and 3.60 m high. A plateau is located at the top of the ramp.
a) Does the ball land on the ramp or plateau? Justify your answer.
b) When it lands, what is its displacement from the launch point.
I know it lands on the incline, but I can't figure out the final components to the displacement or the final displacement itself.
The equations of motion for the ball are
X = 10 cos 50* t = 6.429 t and
Y = 10 sin 50* t - (g/2) t^2 = 7.660 t - 4.90 t^2
t = X/6.429 can be substituted in the second equation to provide a trajectory equation
Y = 7.66 (X/6.429)-4.90(X/6.429)^2
Y = 1.191 X - 0.119 X^2, which is an inverted parabola.
The equation for the ramp slope is
Y' = (3.6/6) X = 0.6 X
If the ball hits the ramp, there is a solution to Y = Y' before the end of the ramp.
1.191 X - 0.119 X^2 = 0.600 X
0.591 X = 0.119 X^2
X = 4.97 meters
This is before the end of the ramp (6.00 m), so the ball hits the ramp rather than the plateau. Use the X value and either equation to solve for Y.
To determine whether the ball lands on the ramp or the plateau, we need to compare the trajectory of the ball with the slope of the ramp.
The trajectory of the ball can be described by the equation Y = 1.191X - 0.119X^2, where X is the horizontal distance traveled by the ball and Y is its vertical position.
The slope of the ramp is given by the equation Y' = 0.6X, where X is the horizontal distance along the ramp and Y' is the vertical position on the ramp.
To find the point where the ball lands, we need to find the intersection of the ball's trajectory and the slope of the ramp. This occurs when Y = Y':
1.191X - 0.119X^2 = 0.6X
Simplifying the equation, we get:
0.591X = 0.119X^2
Dividing both sides by X, we have:
0.591 = 0.119X
Solving for X, we find X = 4.97 meters.
Since this X value is less than the length of the ramp (6.00 m), the ball hits the ramp rather than the plateau.
To find the vertical position (Y) at this point, we can substitute X = 4.97 into either equation. Let's use the equation Y = 1.191X - 0.119X^2:
Y = 1.191(4.97) - 0.119(4.97)^2
Y = 5.91 - 2.96
Y = 2.95 meters
Therefore, the ball lands on the ramp, approximately 4.97 meters from the launch point, with a vertical displacement of 2.95 meters.