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 the value found in part A, at what horizontal distance from the left bank did he land?
To find the horizontal distance from the left bank where the professor landed, we need to calculate the time it takes for the professor to reach the far bank, and then use that time to find the horizontal distance using his speed.
First, let's find the time it takes for the professor to reach the far bank. We can use the kinematic equation for horizontal motion:
x = v0x * t
where x is the horizontal distance, v0x is the horizontal component of velocity, and t is the time.
In this case, the horizontal component of velocity is half the value found in part A, so v0x = v0/2.
The horizontal distance is given as 40.0 m, so we have:
40.0 m = (v0/2) * t
Now, let's find the time using the kinematic equation for vertical motion:
y = v0y * t + (1/2) * a * t^2
where y is the vertical displacement, v0y is the vertical component of velocity, a is the acceleration, and t is the time.
The vertical displacement is 15.0 m, the initial vertical velocity is zero, and the acceleration is due to gravity, -9.8 m/s^2. So we have:
15.0 m = 0 * t + (1/2) * (-9.8 m/s^2) * t^2
Simplifying, we get:
15.0 m = -4.9 m/s^2 * t^2
Rearranging, we have:
t^2 = -15.0 m / (-4.9 m/s^2)
t^2 = 3.06 s
Taking the square root of both sides, we find:
t = 1.75 s
Now, substitute this value of t into the equation for the horizontal distance:
40.0 m = (v0/2) * 1.75 s
Solving for v0, we have:
v0 = (40.0 m) / (0.5 * 1.75 s)
v0 = 45.7 m/s
Therefore, if his speed was only half the value found in part A, he would land at a horizontal distance of 45.7 m from the left bank.