The current world record motorcycle jump is 77.0 meters set by Jason Renie. Assume that he left the take off ramp at 12.0 degrees above horizontal and that the take off and landing heights are the same. Neglecting air resistance,deteremine his take-off speed
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To determine the take-off speed of Jason Renie during the motorcycle jump, we can use the principles of projectile motion. In this case, we'll need to consider the initial vertical velocity and the take-off angle.
First, let's break down the initial velocity into its horizontal and vertical components:
V₀x = V₀ * cos(θ)
V₀y = V₀ * sin(θ)
Here, V₀ is the initial speed or take-off speed, θ is the take-off angle (12.0 degrees in this case), V₀x is the horizontal component of the initial velocity, and V₀y is the vertical component of the initial velocity.
Since the take-off and landing heights are the same, the vertical displacement (Δy) is zero:
Δy = V₀y * t + (1/2) * g * t²
0 = V₀ * sin(12) * t + (1/2) * 9.8 * t²
From the time of flight equation, we know that the total time of the jump is given by:
t = 2 * (V₀ * sin(θ)) / g
Substituting this expression for t into the vertical displacement equation:
0 = (V₀ * sin(12) * 2 * (V₀ * sin(12) / 9.8) / g) + (1/2) * 9.8 * (2 * (V₀ * sin(12) / 9.8) / g)²
Simplifying the equation and solving for V₀:
0 = 2 * (V₀ * sin(12))² / g - (2 * (V₀ * sin(12))² / g) + (1/2) * (2 * (V₀ * sin(12)) / g)²
Combining like terms:
0 = (2 * (V₀ * sin(12))² / g) - (2 * (V₀ * sin(12))² / g) + (1/2) * ((2 * V₀ * sin(12)) / g)²
Now, we can solve this quadratic equation to find the value of V₀.
D = (V^2/g)sin(2A)
2A = 24 degrees. D = 77.0 m
solve for V
He could have gone farther with a steeper angle. Maybe it would have been harder to land.