(a) To find the horizontal distance from the launch point to the landing ramp, we can use the kinematic equation:
y = xtanθ - (gx²) / (2v₀²cos²θ)
Where:
y = 25.0 ft (height difference between launch ramp and landing ramp)
x = horizontal distance we want to find
θ = 30.0o
g = acceleration due to gravity = 32.2 ft/s²
v₀ = initial velocity = 70.0 mph = 102.7 ft/s
Plugging in the values:
25.0 = xtan(30.0) - (32.2x²) / (2(102.7)²cos²(30.0))
25.0 = 0.5774x - 0.0247x²
Rearranging and solving for x, we get:
0.0247x² - 0.5774x + 25.0 = 0
This is a quadratic equation that can be solved using the quadratic formula. The solutions for x will give us the two possible horizontal distances from the launch point to the landing ramp.
(b) To find the angle of the landing ramp, we want it to match the direction of your velocity vector when you touch down. This means the angle of the landing ramp will be 30.0o as well. This ensures that your velocity vector is parallel to the ramp when you land, allowing for a smooth landing.
A Motorcycle Jump. You are planning to make a jump with your motorcycle by driving over a ramp that will launch you at an angle of 30.0o with respect to the horizontal. The front edge of the ramp on which you are supposed to land, however, is 25.0 ft lower than the edge of the launch ramp (i.e., your launch height).
(a) Assuming a launch speed of 70.0 mph, at what horizontal distance from your launch point should the landing ramp be placed?
(b) In order to land smoothly, the angle of the landing ramp should match the direction of your velocity vector when you touch down. What should be the angle of the landing ramp?
1 answer