My Answere to part B should be 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.
I use the following website to solve quadratic equations.
http://www.math.com/students/calculators...
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!