Problem 4: A rocket starts from rest at the origin. The rocket accelerates up an inclined plane of with an angle = 35.0", a length L = 200 m, and an acceleration a = 1.25 m/sec². At a time t the rocket reaches the top of the inclined plane at the location (x, y) with a speed v. See the figure below. a) What is x? Answer: 163.8 m. b) What is y? Answer: 114.7 m. c) What is v, the x-component of the rocket at the top of the inclined plane? Answer: 18.3 m/sec. d) What is vy, the y-component of the rocket at the top of the inclined plane? e) Answer: 12.8 m/sec. y t=0 Xo = 0 Yo=0 y t x =? y =? Vy =? Vy =? /a= 1.25 m/sec² L = 200 m X Vo = 0 Let's now re-define the time to t= 0. At this instant, the acceleration of the rocket shuts off and now we have the rocket in projectile motion. The final properties of the rocket from the rocket at the top of the incline now make the initial properties for the projectile motion. See the figure below. f) What is ymax, the highest elevation of the rocket above the ground? Answer: 123.1 m. g) What is t, the time at which the rocket hits the ground? Answer: 6.3 sec. What is x, the x-position at which the rocket hits the ground? Answer: 279.1 m. h) Re-define as t = 0 xo Yo Vox Voy - LIE |Ymax =? t =? y=0x=? X

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Chapter1: Physics And Measurement
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Problem 1.10OQ: A calculator displays a result as 1.365 218 0 X 107 kg. The estimated uncertainty in the result is...
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Problem 4:
A rocket starts from rest at the origin. The rocket accelerates up an inclined plane of with an
angle 0 = 35.0°, a length L = 200 m, and an acceleration a = 1.25 m/sec². At a time t the
rocket reaches the top of the inclined plane at the location (x, y) with a speed v. See the figure
below.
a) What is x? Answer: 163.8 m.
b) What is y? Answer: 114.7 m.
c)
What is v₂, the x-component of the rocket at the top of the inclined plane?
Answer: 18.3 m/sec.
d) What is vy, the y -component of the rocket at the top of the inclined plane?
e) Answer: 12.8 m/sec.
y
y
t = 0
Xo = 0
Yo=0
0
t
x = ? y =?
Vx =? Vy = ?
a = 1.25 m/sec²
L = 200 m
Vo = 0
Let's now re-define the time to t = 0. At this instant, the acceleration of the rocket shuts off
and now we have the rocket in projectile motion. The final properties of the rocket from the
rocket at the top of the incline now make the initial properties for the projectile motion.
See the figure below.
f) What is ymax, the highest elevation of the rocket above the ground? Answer: 123.1 m.
g) What is t, the time at which the rocket hits the ground? Answer: 6.3 sec.
h) What
is x, the x-position at which the rocket hits the ground? Answer: 279.1 m.
x
Re-define as t = 0
xo Yo
LIE
Vox Poyr
|Ymax =?
t=?
y = 0 x =?
x
Transcribed Image Text:Problem 4: A rocket starts from rest at the origin. The rocket accelerates up an inclined plane of with an angle 0 = 35.0°, a length L = 200 m, and an acceleration a = 1.25 m/sec². At a time t the rocket reaches the top of the inclined plane at the location (x, y) with a speed v. See the figure below. a) What is x? Answer: 163.8 m. b) What is y? Answer: 114.7 m. c) What is v₂, the x-component of the rocket at the top of the inclined plane? Answer: 18.3 m/sec. d) What is vy, the y -component of the rocket at the top of the inclined plane? e) Answer: 12.8 m/sec. y y t = 0 Xo = 0 Yo=0 0 t x = ? y =? Vx =? Vy = ? a = 1.25 m/sec² L = 200 m Vo = 0 Let's now re-define the time to t = 0. At this instant, the acceleration of the rocket shuts off and now we have the rocket in projectile motion. The final properties of the rocket from the rocket at the top of the incline now make the initial properties for the projectile motion. See the figure below. f) What is ymax, the highest elevation of the rocket above the ground? Answer: 123.1 m. g) What is t, the time at which the rocket hits the ground? Answer: 6.3 sec. h) What is x, the x-position at which the rocket hits the ground? Answer: 279.1 m. x Re-define as t = 0 xo Yo LIE Vox Poyr |Ymax =? t=? y = 0 x =? x
-b+√b²-4ac
2a
x = x₂ + V₁xt + ²axt², V₁₁x = V₁x + axt, (v₁₁x)² = (v₁x)² + 2axx, t ==
у = V₁ + V₁¸yt + ² α¸t², Vf₁y = V₁y+āÿt, (Vƒ‚y)² = (v₁y)² + 2a,ÿ
Transcribed Image Text:-b+√b²-4ac 2a x = x₂ + V₁xt + ²axt², V₁₁x = V₁x + axt, (v₁₁x)² = (v₁x)² + 2axx, t == у = V₁ + V₁¸yt + ² α¸t², Vf₁y = V₁y+āÿt, (Vƒ‚y)² = (v₁y)² + 2a,ÿ
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