Asked by Andy
A rocket is launched into the sky on a windy day. The rocket has a vertical velocity of 15 m/s. There is a strong wind blowing east to west at 35 m/s. How far from the start point is the rocket after 60 seconds?
Answers
Answered by
Steve
vertical travel:
y = 15t
horizontal travel:
x = 35t
actual distance:
d = √(x^2+y^2)
= √(15t)^2 + (35t)^2)
= 5√58 t
At t=60, d=2285 m
Assuming the rocket does not run out of fuel and go ballistic (literally)
y = 15t
horizontal travel:
x = 35t
actual distance:
d = √(x^2+y^2)
= √(15t)^2 + (35t)^2)
= 5√58 t
At t=60, d=2285 m
Assuming the rocket does not run out of fuel and go ballistic (literally)
Answered by
qazi
vertical travel:
y = 15t
horizontal travel:
x = 35t
actual distance:
d = √(x^2+y^2)
= √(15t)^2 + (35t)^2)
= 5√58 t
At t=60, d=2285 m
y = 15t
horizontal travel:
x = 35t
actual distance:
d = √(x^2+y^2)
= √(15t)^2 + (35t)^2)
= 5√58 t
At t=60, d=2285 m
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