Height of a rocket at a certain time

AI Thread Summary
The discussion revolves around calculating the height of a rocket when a discarded canister hits the launch pad. The rocket accelerates at 3.20 m/s² and reaches a speed of 40.8 m/s at 260 m altitude. After the canister is released, it ascends to a maximum height before falling back, with gravity as the only acting force. The user initially miscalculated the final height of the rocket, arriving at 715 m, and sought clarification on the need to account for multiple time intervals in the calculations. The solution involves determining the time to reach 260 m, the time to the canister's maximum height, and the time of free fall, which collectively inform the final height of the rocket.
moy13
Messages
13
Reaction score
0

Homework Statement



During launches, rockets often discard unneeded parts. A certain rocket starts from rest on the launch pad and accelerates upward at a steady 3.20 m/s^2. When it is 260 m above the launch pad, it discards a used fuel canister by simply disconnecting it. Once it is disconnected, the only force acting on the canister is gravity (air resistance can be ignored).

How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration?

Homework Equations



1. v^2 = v(0)^2 - 2g(y - y(0))
2. y = y(0) + v(0)t - (1/2)at^2
3. v = v(0) - gt

The Attempt at a Solution



first i wanted to find the speed of the rocket at 260m by using equation 1. and solving for v. I attained 40.8m/s.

then I used this 40.8m/s as the initial velocity of the canister going up and v = 0 (speed of the canister at its maximum height) as the final velocity to find the initial height of the canister before it begins to drop. I attained 345m.

once i found this i wanted to find the final velocity of the canister before it hit the pad and used equation 1 again. I attained v = 82.2m/s.

after, i found the time it took the canister to hit the ground by using equation 3. I attained 8.39s.

finally, i used this time interval along with v(0) = 40.8m/s, y(0) = 260m, a = 3.2m/s^2 in equation 2. I attained 715m which is wrong according to the course website.

can anybody help me?
 
Last edited:
Physics news on Phys.org
By your method there are 3 time intervals that need to be determined.

1. time to 260 m
2. time to canister max
3. time to free fall.

40.8 m/s / 3.2 yields your first time.
40.8 m/s / 9.8 yields your second time
345m = 1/2*9.8*t2 yields your third.

Add all 3 together and that put back into 1/2*(3.2)*t2 should yield your rocket height right?

Alternatively you can plug the 40.8 and the 260 into the initial conditions relating time, distance,and velocity and solve the quadratic
 
wow! Thank you so much LowlyPion. I did not notice how there were more than one time intervals.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top