A physics professor did daredevil stunts in his spare time H

A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump on a motorcycle (the figure (Figure 1)). The take off ramp was inclined at 53.0 degree, 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. What should his speed have been at the top of the ramp to have just made it to the edge of the far bank? It is speed was only half the value found in part A, at what horizontal distance from the left bank did he land?

Solution

1st let’s determine the horizontal and vertical components of the initial velocity.

Horizontal = v * cos 53, Vertical = v * sin 53

To reach the edge of the far bank, the professor must move 40 meters horizontally in the same time as he moves 15 meters downward. Let’s use the following equation to determine the time for the professor to move 40 meters horizontally.

Horizontal distance = v * cos * t
40 = v * cos 53 * t
Eq#1: t = 40 ÷ v * cos 53

Let’s use the following equation to determine and the professor’s vertical displacement in amount of time to determine the professor’s initial velocity.

d = vi * t + ½ * a * t^2, d = -15, vi = v * sin 53, a = -9.8
d is the vertical displacement = Final height – Initial height = 0 – 15 = -15 m

Eq#2: -15 = v * sin 53 * t – ½ * 9.8 * t^2
Substitute (40 ÷ v * cos 53) for t

-15 = v * sin 53 * (40 ÷ v * cos 53) – 4.9 * (40 ÷ v * cos 53)^2

-15 = tan 53 * 40 – 4.9 * (40 ÷ v * cos 53)^2

-15 – tan 53 * 40 = -4.9 * (40 ÷ v * cos 53)^2
Divide both sides by -4.9

(-15 – tan 53 * 40) ÷ -4.9 = (40 ÷ v * cos 53)^2
13.89424344 = 1600 ÷ (v * cos 53)^2
(v * cos 53)^2 * 13.89424344 = 1600
v^2 = 1600 ÷ (cos^2 53 * 13.89424344)

Multiply both sides by (v * cos 53)^2

(v * cos 53)^2 * [(-15 – tan 53 * 40) ÷ -4.9] = 1600

Divide both sides by cos^2 15 * [(-15 – tan 53 * 40) ÷ -4.9]

v^2 = 1600 ÷ (cos^2 15 * [(-15 – tan 53 * 40) ÷ -4.9]) = 317.9501462
v = [1600 ÷ (cos^2 15 * [(-15 – tan 53 * 40) ÷ -4.9])] = 317.9501462 =17.83115661 m/s
This is approximately 17.831 m/s

Let’s substitute this number into Eq#1 and determine the time.
t = 40 ÷ 17.83115661* cos 53 = 3.727498282 seconds

Let’s substitute the velocity and time into Eq#2.
-15 = 17.83115661* sin 53 * 3.727498282 – ½ * 9.8 * 3.727498282^2
53.08179285 – 68.08179287 = -15
This proves that the initial velocity is correct!

If his speed was only half the value found in A, where did he land?
v = ½ * 17.83115661 = 8.915578305 m/s

Horizontal = v * cos 53, Vertical = v * sin 53
Let’s determine the initial horizontal and vertical velocity.
Horizontal = 8.915578305 * cos 53
Vertical = 8.915578305 * sin 53

Let’s use the equation below to determine the time for the professor to move from the ramp to the river,

d = v * sin 53 * t – ½ * 9.8 * t^2
d = -100
-100 = 8.915578305 * sin 53 * t – 4.9 * t^2
4.9 * t – 8.915578305 * sin 53 * t – 100 = 0
Let’s solve this quadratic equation to determine the time.

solve quadratic equations.

t = 5.302154267371364 seconds

Let’s use this time and the initial horizontal velocity to determine the horizontal distance the professor travels.

Initial horizontal velocity = 8.915578305 * cos 53
Horizontal distance = 8.915578305 * cos 53 * 5.302154267371364 = 28.44886229 meters
Since this is less than 40 meters, the professor lands in the river!