A daredevil's last stunt was to attempt to jump across a river on a motorcycle. The takeoff ramp was inclined at 53.0 degrees, the river was 40.0 m wide, and the far bank was 15.0 m lower than the top of the ramp. The river itself was 100 m below the ramp. Ignoring air resistance, what speed was needed at the top of the ramp for the daredevil to have just made it to the edge of the far bank?
10 points to the person who not only tells me the answer, but how to solve it.
Thanks in advance. =)
10 points to the person who not only tells me the answer, but how to solve it.
Thanks in advance. =)
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Let V be the takeoff velocity.
v0v = V * sin(53)
v0h = V * cos(53)
p(t) = 0.5at^2 + v0 * t + p0
The horizontal motion, using the end of the ramp as position 0 and going across the river is the positive direction:
40 = 0.5 * 0 * t^2 + v0h * t + 0
40 = v0h * t
40 / v0h = t
The vertical motion, using the top of the ramp as position 0 and -9.8 m/s^2 as the acceleration due to gravity:
-15 = 0.5 * (-9.8) * t^2 + v0v * t + 0
-15 = (-4.9) * t^2 + v0v * t
Substitute for t:
-15 = (-4.9) * (40 / v0h)^2 + v0v * (40 / v0h)
Substitute for v0h and v0v:
-15 = (-4.9) * (40 / (V * cos(53)))^2 + V * sin(53) * (40 / (V * cos(53)))
-15 = (-4.9) * (40 / (V * cos(53)))^2 + sin(53) * (40 / cos(53))
Solve for V.
v0v = V * sin(53)
v0h = V * cos(53)
p(t) = 0.5at^2 + v0 * t + p0
The horizontal motion, using the end of the ramp as position 0 and going across the river is the positive direction:
40 = 0.5 * 0 * t^2 + v0h * t + 0
40 = v0h * t
40 / v0h = t
The vertical motion, using the top of the ramp as position 0 and -9.8 m/s^2 as the acceleration due to gravity:
-15 = 0.5 * (-9.8) * t^2 + v0v * t + 0
-15 = (-4.9) * t^2 + v0v * t
Substitute for t:
-15 = (-4.9) * (40 / v0h)^2 + v0v * (40 / v0h)
Substitute for v0h and v0v:
-15 = (-4.9) * (40 / (V * cos(53)))^2 + V * sin(53) * (40 / (V * cos(53)))
-15 = (-4.9) * (40 / (V * cos(53)))^2 + sin(53) * (40 / cos(53))
Solve for V.