5. A test rocket is launched vertically from ground level (y = 0 m), at time t

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 5 A test rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 49 m and acquired a velocity of 30m/s  What is the maximum height that the rocket will reach?
 6 Aracquetball strikes aWall with aSpeed of 30 m/s and rebounds with a speed of 26m/s  The collision takes20milisec What is theAverage cceleration of the ball during the collision?
May 17th, 2015

Hello! :)

5.

We know that the rocket has already risen to 49 meters, so that will be added to the height that we calculate.  We will assume that 49 meters is the initial point where the rocket starts at 30m/s.  The main equation below will be used to calculate the maximum height that the rocket will reach.

ymax = {((v2)^2 - (v1)^2)/(2g)} + 49

ymax = max height

v2 = rocket's velocity when it hits maximum elevation = 0m/s

v1 = initial velocity starting at 49m height = 30m/s

g = -9.8 m/(s^2)

ymax = {((0^2) - (30^2))/(2*-9.8)} + 49

ymax = {200/19.6) + 49

ymax = 10.2 + 49

ymax = 59.2 meters

(The rocket travels an extra 10.2 meters once the engine burns out)

6.

Use the equation below to solve for average acceleration:

Average acceleration = (change in velocity) / (time)

v1 = 30m/s

v2 = -26m/s

change in velocity = v2 - v1

Convert time from milliseconds (ms) to seconds (s)

t = 20ms * (1s/1000ms) = 0.02 seconds

Average acceleration = (26-30)/.02 = -43.3 m/s

Let me know if you have any questions!  Hope this helps :)

May 17th, 2015

both wrong...... i dont have any of those answers in my options

May 17th, 2015

hello

May 17th, 2015

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May 17th, 2015
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May 17th, 2015
Dec 9th, 2016
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