Long Problems P1 A rocket is launched straight up at constant acceleration a-4.2

Question

Long Problems P1 A rocket is launched straight up at constant acceleration a-4.2 m/s2. Four seconds after liftoff, a bolt falls off the side of the rocket. Please answer each of the following questions (a) What is the velocity of the rocket when the bolt falls off? (b) How long after it drops off does it take for the bolt to hit the ground? P2 A skateboarder leaves a ramp at a height of 1 m at 7 m/s, at 30° above the horizontal. Disregard air resistance. Please answer each of the following (a) How long does she stay in the air? (b) How far from the base of the ramp does she hit the ground? (c) What is her velocity when she hits the ground? P3 Ships A and B leave port at the same time. For the next two hours ship A travels at 20mph in a direction 30° west of north while ship B travels 20° east of north at 25 mph (a) What is the distance between the two ships two hours after they depart? (b) What is the velocity of ship A as seen by ship B P4 At t- 0, an object is located at x -10 m and travelling in the *x direction at 15 m/s. From then on, it accelerates according to the function a(t) btc, where b-2 m/s3, and c 6m/s2. Please answer each of the following questions a) What is the velocity of the ball att 5 s? b) What is the maximum velocity of the ball?

Explanation / Answer

P1 :

a)

consider the motion along the Y-direction for the rocket :

a = acceleration of the rocket = 4.2 m/s2

voy = initial velocity = 0 m/s

vfy = final velocity = ?

t = time of travel = 4 sec

Using the equation

vfy = voy + at

vfy = 0 + (4.2) (4)

vfy = 16.8 m/s

b)

consider the motion of rocket for first 4.2 sec along the Y-direction

ar = acceleration of the rocket = 4.2 m/s2

voy = initial velocity = 0 m/s

Yo = initial position = 0 m

t = time of travel = 4 sec

Y = final position after time "t"

Y = Yo + voy t + (0.5) ar t2

Y = 0 + 0 (4) + (0.5) (4.2) (4)2

Y = 33.6 m

consider the motion of bolt after droping off along the Y-direction

Voy = initial velocity In Y-direction = 16.8 m/s

a = acceleration = - 9.8

Yo = initial position = 33.6 m

Y = final position = 0 m

t' = time taken by the bolt to hit the ground

using the equation

Y = Yo + Voy t' + (0.5) at'2

0 = 33.6 + (16.8) t' + (0.5) (- 9.8) t'2

t = 4.84 sec

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