engine of a toy rocket supplies an average of 38.0 m/s^2 to the rocket for an interval of .80 s theres more?

Question

If the toy rocket is launched vertically, how high does it rise in the interval
How fast is the rocket moving at the end of .80 s
What altitude does the rocket reach before falling back to earth
How long does it take to reach this altitude

Answer 1

d = .5at² = .5*38*.8² = 12.16 meters
v = at = 38*.8 = 30.4 m/s
At the top of it's trajectory, it's velocity is 0 and
v² = v0² - 2gh where v0 is initial velocity, g is gravity, and h is height. So
h = 30.4²/(2*9.8) = 23.576 meters above where the motor burned out (12.16 m) Therefore the actual height is
23.576 + 12.16 = 35.736 m

You gotta learn what those equations **mean** Dewd. And when and how to use them ☺


Doug

Answer 2

Woo. Too complicated for me. This is where you need calculus. The speed changes from 0 to 38 mps in 1 second, but I don't think that the acceleration would graph to a straight line. The mass of the fuel is decreasing as it burns, reaching zero in .8 seconds, so, the max acceleration is reached at .8 seconds. This is not quit 38 mps since the burn did not last 1 full second.

Speed at .8 second? Let's say 37 mps? G=32 fps^s
37 mps = 121.39108 fps (A)
H= (A) + (A-G)
H= 121.391081121.39108
89.39108189.39108210.78216
57.39108157.39108268.17324
25.39108125.39108293.56432
9.391080.534.78216328.34648
1.391080.2536.17324364.51972
0.391080.0312536.56432401.08404

So it would fly about 5 seconds to about 400 feet.
But this is all just an estimate. There is no drag mentioned and that is the biggest limiter for a rocket.

;-D Is it carrying a camera? A parachute?