Best answer:
The path of the rocket is
y=-4.9t^2+39.2t+h,
where y is the height of the rocket at time t s
measured from the ground. h is the height of
the cliff.
(a) After 10 s from start on top of the cliff, the
rocket hits the ground=>
0=-4.9(10^2)+39.2(10)+h=>
h=98 m
(c)...
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Best answer: The path of the rocket is
y=-4.9t^2+39.2t+h,
where y is the height of the rocket at time t s
measured from the ground. h is the height of
the cliff.
(a) After 10 s from start on top of the cliff, the
rocket hits the ground=>
0=-4.9(10^2)+39.2(10)+h=>
h=98 m
(c) y=-4.9t^2+39.2t+98=>
y'=-9.8t+39.2=0=>
t=4 s is the time needed for the rocket to reach
the max. height after start.
(b) The max. height
H=-4.9(4^2)+39.2(4)+98=>
H=176.4 m
(d) y'=-9.8(10)+39.2=-58.8 m/s
is the speed of the rocket when
hitting the ground.
(e) When the rocket passes by
the top of the cliff,
y=98=>
98=-4.9t^2+39.2t+98=>
t(-4.9t+39.2)=0=>
t=0 or t=39.2/4.9=8 s (when the rocket
bypasses it when coming down).
Ans. 8 s.
4 answers
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5 days ago