A rocket moves upward, starting from rest with an acceleration of 28.8 m/s2 for 4.13 s. It runs out of fuel at the end of the 4.13 s but does not stop. How high does it rise above the ground?
physics - bobpursley, Sunday, December 5, 2010 at 6:27pm
But thrust*distancefiring=energy put into the rocket. How high?
energy put into rocket=mgh
solve for h.
now distance firing: d=1/2 a t^2
physics - drwls, Sunday, December 5, 2010 at 6:30pm
Compute the distance that it rises while thrusting for time t. Call it X1
X1 = (1/2) a t^2
Also compute the velocity when acceleration stops. Call it Vmax.
Vmax = a t
Compute how much farther the rocket travels up until the maximum height (and zero velocity) is reached. Call that X2.
Maximum height is reached after additional time interval
t' = Vmax.g
X2 = Vmax*t' - (g/2)t'^2
The answer is X1 + X2.
physics - tchrwill, Tuesday, February 15, 2011 at 4:58pm
Liftoff to burnout in 4.13 sec.
Burnout velocity Vbo = 28.8(4.13) = 123.84m/s.
Burnout altitude h1 = at^2/2 - gt^2/2 = 19(4.13)^2/2 = 162m.
From Vbo to V = 0, Vf = Vbo - gt = 0 = 123.84 - 9.8t or t = 12.636 sec.
From Vbo to V = 0, h2 = 123.84(12.636) - 9.8(12.636)^2/2 = 782m.
Total height H = h1 + h2.