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Question 1087259:
A toy rocket is launched from the top of a building 67 feet tall at an initial velocity of 237 feet per second.
a) Give the function that describes the height of the rocket in terms of time t.
b) Determine the time at which the rocket reaches its maximum height, and the maximum height in feet.
c) For what time interval will the rocket be more than 6969 feet above ground level?
d) After how many seconds will it hit the ground?
Found 2 solutions by Alan3354, jorel1380:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! A toy rocket is launched from the top of a building 67 feet tall at an initial velocity of 237 feet per second.
a) Give the function that describes the height of the rocket in terms of time t.
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You should provide that.
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w/o the function, nothing else can be done.
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b) Determine the time at which the rocket reaches its maximum height, and the maximum height in feet.
c) For what time interval will the rocket be more than 6969 feet above ground level?
d) After how many seconds will it hit the ground?
Answer by jorel1380(3719)  (Show Source):
You can put this solution on YOUR website! The equation for this rockets height at any time t (in seconds) would be:
a) h(t)=-4.9tē+237t+67
b) the maximum height of a projectile would be reached at a time of h(t)= atē+bt+c is given by the value -b/2a, which, in this case, would be -237/-9.8=24.184 seconds after launch.
The maximum height, given by the above equation, would be 2932.765 ft.
c)To find the time interval for when the rocket is above 69 ft., we use
69=-4.9tē+237t+67
4.9tē-237t+2=0
Using the quadratic formula, we get two roots of 48.3589066473 and 0.0084402914293 seconds. This gives us about 48 seconds where the rocket is higher than 69 ft.
d)h(t)=0=-4.9tē+237t+67
4.9tē-237t-67=0
Using the quadratic formula again, we get a positive value for t of 48.6484140538 seconds total flight time. ☺☺☺☺
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