Asked by nicky
A rocket moves upward, starting from rest with an acceleration of 25.0 m/s2 for 3.77 s. It runs out of fuel at the end of the 3.77 s but does not stop. How high does it rise above the ground
I`ve been sick last week, not quite sure where to even start can you give me a formula or some guidance thanks!
I`ve been sick last week, not quite sure where to even start can you give me a formula or some guidance thanks!
Answers
Answered by
Damon
power stage
v = Vi + a t
v = 0 + 25 * 3.77
v = 94.25 m/s when fuel runs out
h = 0 + 0 t + (1/2)(25)(3.77)^2
h = 177.7 meters high when fuel runs out
Coasting stage, Hi = 177.7 meters and Vi = 94.25 m/s
then
at top v = 0
0 = 94.25 - 9.81 t
t = 9.61 seconds coasting up
h = Hi + Vi t - 4.9 t^2
h = 177.7 + 94.25(9.61)-4.9(9.61)^2
v = Vi + a t
v = 0 + 25 * 3.77
v = 94.25 m/s when fuel runs out
h = 0 + 0 t + (1/2)(25)(3.77)^2
h = 177.7 meters high when fuel runs out
Coasting stage, Hi = 177.7 meters and Vi = 94.25 m/s
then
at top v = 0
0 = 94.25 - 9.81 t
t = 9.61 seconds coasting up
h = Hi + Vi t - 4.9 t^2
h = 177.7 + 94.25(9.61)-4.9(9.61)^2
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