a o https session. mastering myc emView? assignment Problem 54476043 MasteringPh

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a o https session. mastering myc emView? assignment Problem 54476043 MasteringPhysics: Assignment 16 ngphysics.com PHYSICS 201 SECT 01 FALL 2015 Assignment 16 Problem 8.23 Problem 8.23 Part A Sam (70 kg) takes off up a 50-m-high, 10 frictionless slope on How far does Sam land from the base of the cliff? his jet-powered skis. The skis have a thrust of 180 N. He keeps Express your answer to two significant figures and include the appropriate units. his skis tilted at 10 after becoming airborne, as shown in the figure(Figure 1) Ar 149.55 m Submit My Answers Give Up Incorrect; Try Again; 4 attempts remaining Figure 1 of 1 10° Start 50 m 10° C Reade Help Close Resources previous l 2 of 3 l next ovide Feedback Continue

Explanation / Answer

.
Ramp length d = 50/sin(theta)
Net acceleration a = F/m - gsin(theta)
Launch velocity v = sqrt(2ad)
Now it's just a simple trajectory analysis: v and theta are known, and there is a height change of -50 m. The following equations assume launch height = 0 and final height yt = -50 m; the same situation but shifted vertically:

V0y = 4.0933 m/s
S0y = 50 m (initial vertical displacement)

S_final_y = zero (ground level)

The vertical component of his acceleration is (-weight + upward component of thrust) / mass
Which is (-mg + 200*sin10) / m = -700.27 newtons / (70 kg) = -10.00 m/s^2 (note that he accelerates toward the ground at a rate slightly less than one gee due to his rockets)

using s = s0 + Vo * t + (1/2) * a * t^2 we get

0 = 50 + 4.0933 * t + 0.5 *(-10.00) * t^2
0 = 50 + 4.0933 t - 5 t^2

Solving the quadratic gives

3.59 seconds

so he is airborne for 3.59seconds

Now we have to repeat the process for his horizontal motion:

S0_x = 0
V0_x = 23.572

A0_x = 200*cos(10) / 70 = 2.8137 m/s^2


S_final_x = S0_x + V0_x * t + (1/2)*A0_x*t^2


S_final_x =0 + 23.572 * 3.59 + (1/2)*2.8137*3.59^2
= 102.75 meters

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