Answer:
At highest point:
y1 = 10.4 ft
v1 = (26.5*i + 0*j) ft/s
When he lands:
x2 = 31.5 ft (distance he travels)
t2 = 1.19 s
V2 = (26.5*i - 25.9*j) ft/s
a2 = -44.3°
Explanation:
Since he let go of the tow rope upon leaving the ramp he is in free fall from that moment on. In free fall he is affected only by the acceleration of gravity. Gravity has a vertical component only, so the movement will be at constant acceleration in the vertical component and at constant speed in the horizontal component.
20 mi / h = 29.3 ft/s
If the ramp has an angle of 25 degrees, the speed is
v0 = (29.3 * cos(25) * i + 29.3 * sin(25) * j) ft/s
v0 = (26.5*i + 12.4*j) ft/s
I set up the coordinate system with the origin at the base of the ramp under its end, so:
R0 = (0*i + 8*j) ft
The equation for the horizontal position is:
X(t) = X0 + Vx0 * t
The equation for horizontal speed is:
Vx(t) = Vx0
The equation for vertical position is:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
The equation for vertical speed is:
Vy(t) = Vy0 + a * t
In this frame of reference a is the acceleration of gravity and its values is -32.2 ft/s^2.
In the heighest point of the trajectory the vertical speed will be zero because that is the point where it transitions form going upwards (positive vertical speed) to going down (negative vertical speed), and it crosses zero.
0 = Vy0 + a * t1
a * t1 = -Vy0
t1 = -Vy0 / a
t1 = -12.4 / -32.2 = 0.38 s
y1 = y(0.38) = 8 + 12.4 * 0.38 + 1/2 * (-32.2) * (0.38)^2 = 10.4 ft
The velocity at that moment will be:
v1 = (26.5*i + 0*j) ft/s
When he lands in the water his height is zero.
0 = 8 + 12.4 * t2 + 1/2 * (-32.2) * t2^2
-16.1 * t2^2 + 12.4 * t2 + 8 = 0
Solving this equation electronically:
t2 = 1.19 s
Replacing this time on the position equation:
X(1.19) = 26.5 * 1.19 = 31.5 ft
The speed is:
Vx2 = 26.5 ft/s
Vy2 = 12.4 - 32.2 * 1.19 = -25.9 ft/s
V2 = (26.5*i - 25.9*j) ft/s
a2 = arctg(-25.9 / 26.5) = -44.3
