SOLUTION: the formula h=vt-16t^2 gives the height h in feet of an object after t sec. Here v is the initial velocity of the object expressed in feet per second. suppose a toy rocket was laun

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Question 262848: the formula h=vt-16t^2 gives the height h in feet of an object after t sec. Here v is the initial velocity of the object expressed in feet per second. suppose a toy rocket was launched from the ground straight up with an initial velocity of 64ft/sec. How many seconds after launch was the rocket 48 ft above the ground.
ok so here is what i have so far:
using the formula and what is given i have 48=64t-16t^2 so i need to make it a quadratic like this 64t-16t^2-48=0 then multiply by -1 and put it in the correct order i get:
16t^2-64t+48=0
16(t^2-4t+3)=0
16(t-3)(t-1)
so i get two positive answers. when i put t=1 into the formula like this:
48=64(1)-16(1^2) I get
48=48
but how can something launched at 64ft per second only be 48 feet up after 1 second? is my answer wrong or am i missing something in the physics of the problem?
i hope my notation is clear. any help you can give would be appreciated.
Found 2 solutions by dabanfield, stanbon:
Answer by dabanfield(803) (Show Source):
You can put this solution on YOUR website!
What you have looks good. Even though the intitial velocity is 64ft/sec, after one second the rocket would be only 64 ft high in the absence of gravity. So it makes sense that after one second, with gravity, it would be less than 64 feet high. After two more seconds (that is at t=3) the rocket has reached its high point and is descending. The rocket's path is a parabola and so there are two times when the rocket is at 48 feet, once on the way up and once on the way down again.

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
the formula h=vt-16t^2 gives the height h in feet of an object after t sec. Here v is the initial velocity of the object expressed in feet per second. suppose a toy rocket was launched from the ground straight up with an initial velocity of 64ft/sec. How many seconds after launch was the rocket 48 ft above the ground.
ok so here is what i have so far:
using the formula and what is given i have 48=64t-16t^2 so i need to make it a quadratic like this 64t-16t^2-48=0 then multiply by -1 and put it in the correct order i get:
16t^2-64t+48=0
16(t^2-4t+3)=0
16(t-3)(t-1)
so i get two positive answers. when i put t=1 into the formula like this:
48=64(1)-16(1^2) I get
48=48
but how can something launched at 64ft per second only be 48 feet up after 1 second? is my answer wrong or am i missing something in the physics of the problem?
i hope my notation is clear. any help you can give would be appreciated.
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the formula h=vt-16t^2 gives the height h in feet of an object after t sec. Here v is the initial velocity of the object expressed in feet per second. suppose a toy rocket was launched from the ground straight up with an initial velocity of 64ft/sec. How many seconds after launch was the rocket 48 ft above the ground.
ok so here is what i have so far:
using the formula and what is given i have 48=64t-16t^2 so i need to make it a quadratic like this 64t-16t^2-48=0 then multiply by -1 and put it in the correct order i get:
16t^2-64t+48=0
16(t^2-4t+3)=0
16(t-3)(t-1)
so i get two positive answers. when i put t=1 into the formula like this:
48=64(1)-16(1^2) I get
48=48
but how can something launched at 64ft per second only be 48 feet up after 1 second? is my answer wrong or am i missing something in the physics of the problem?
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Ans: Gravity is pulling the object back to earth at the rate of 16 ft.
per second
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I hope my notation is clear. any help you can give would be appreciated.
the formula h(t) = vt-16t^2 gives the height h in feet of an object after t sec.
How many seconds after launch was the rocket 48 ft above the ground.
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-16t^2 + 64t = 48
-16t^2 + 64t - 48 = 0
-16(t^2 - 4t + 3) = 0
t^2 - 4t + 3 = 0
(t-3)(t-1) = 0
t = 1 second ; t = 3 seconds
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Cheers,
Stan H.