A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle (Figure 1). The takeoff ramp was inclined at 53.0∘, 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. You can ignore air resistance.
If his speed was only half 17.8 m/s, at what horizontal distance from the left bank did he land?
To solve this problem, we can use the equations of motion. Let's first analyze the motion in the x-direction.
Since there is no air resistance, the horizontal component of the velocity remains constant throughout the motion. Let's call it Vx.
We know that the speed of the motorcycle is half of 17.8 m/s, so Vx = 17.8 m/s / 2 = 8.9 m/s.
The horizontal distance the professor will travel in the x-direction can be found using the equation:
x = Vx * t
To find the time of flight, we need to consider the vertical motion. Let's analyze the motion in the y-direction.
We can break down the initial velocity into its vertical and horizontal components. The vertical component of the velocity, Vy, can be found using trigonometry:
Vy = V * sin(θ)
where V is the initial speed of the motorcycle, which is 17.8 m/s, and θ is the angle of the ramp, which is 53.0∘.
Using this, we can find the time of flight, t:
t = (2 * Vy) / g
where g is the acceleration due to gravity, which is approximately 9.8 m/s².
Substituting the values, we have:
t = (2 * 17.8 m/s * sin(53.0∘)) / 9.8 m/s²
After calculating this, we can substitute the value of t into the equation for x to find the horizontal distance from the left bank where he lands.
x = 8.9 m/s * t
Calculating the values, we find that the professor lands approximately 17.9 m from the left bank of the river.