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Answer :
Sure! Let's solve the problem step-by-step:
We are given the height function of a toy rocket launched vertically upward as [tex]\( h(t) = -16t^2 + 128t \)[/tex], where [tex]\( h \)[/tex] is the height in feet and [tex]\( t \)[/tex] is the time in seconds.
### a. How long will it take the rocket to hit its maximum height?
To find the time at which the rocket reaches its maximum height, we need to determine when the velocity is zero. The velocity, [tex]\( v(t) \)[/tex], is the derivative of the height function [tex]\( h(t) \)[/tex]:
[tex]\[ v(t) = \frac{d[h(t)]}{dt} \][/tex]
Calculating the derivative:
[tex]\[ v(t) = \frac{d[-16t^2 + 128t]}{dt} \][/tex]
[tex]\[ v(t) = -32t + 128 \][/tex]
To find the time when the velocity is zero:
[tex]\[ -32t + 128 = 0 \][/tex]
[tex]\[ 32t = 128 \][/tex]
[tex]\[ t = \frac{128}{32} \][/tex]
[tex]\[ t = 4 \][/tex]
So, it takes 4 seconds for the rocket to reach its maximum height.
### b. What is the maximum height?
To find the maximum height, we substitute the time [tex]\( t = 4 \)[/tex] back into the original height function [tex]\( h(t) \)[/tex]:
[tex]\[ h(4) = -16(4)^2 + 128(4) \][/tex]
[tex]\[ h(4) = -16(16) + 128(4) \][/tex]
[tex]\[ h(4) = -256 + 512 \][/tex]
[tex]\[ h(4) = 256 \][/tex]
Thus, the maximum height is 256 feet.
### c. How long did it take for the rocket to reach the ground?
The rocket reaches the ground when the height [tex]\( h(t) = 0 \)[/tex]. We need to solve the equation:
[tex]\[ -16t^2 + 128t = 0 \][/tex]
Factor out the common term [tex]\( t \)[/tex]:
[tex]\[ t(-16t + 128) = 0 \][/tex]
This gives us two solutions:
[tex]\[ t = 0 \][/tex] (the time the rocket is launched) or
[tex]\[ -16t + 128 = 0 \][/tex]
[tex]\[ 16t = 128 \][/tex]
[tex]\[ t = \frac{128}{16} \][/tex]
[tex]\[ t = 8 \][/tex]
So, it takes 8 seconds for the rocket to return to the ground.
### Summary:
- The rocket reaches its maximum height after 4 seconds.
- The maximum height of the rocket is 256 feet.
- The rocket takes 8 seconds to reach the ground.
We are given the height function of a toy rocket launched vertically upward as [tex]\( h(t) = -16t^2 + 128t \)[/tex], where [tex]\( h \)[/tex] is the height in feet and [tex]\( t \)[/tex] is the time in seconds.
### a. How long will it take the rocket to hit its maximum height?
To find the time at which the rocket reaches its maximum height, we need to determine when the velocity is zero. The velocity, [tex]\( v(t) \)[/tex], is the derivative of the height function [tex]\( h(t) \)[/tex]:
[tex]\[ v(t) = \frac{d[h(t)]}{dt} \][/tex]
Calculating the derivative:
[tex]\[ v(t) = \frac{d[-16t^2 + 128t]}{dt} \][/tex]
[tex]\[ v(t) = -32t + 128 \][/tex]
To find the time when the velocity is zero:
[tex]\[ -32t + 128 = 0 \][/tex]
[tex]\[ 32t = 128 \][/tex]
[tex]\[ t = \frac{128}{32} \][/tex]
[tex]\[ t = 4 \][/tex]
So, it takes 4 seconds for the rocket to reach its maximum height.
### b. What is the maximum height?
To find the maximum height, we substitute the time [tex]\( t = 4 \)[/tex] back into the original height function [tex]\( h(t) \)[/tex]:
[tex]\[ h(4) = -16(4)^2 + 128(4) \][/tex]
[tex]\[ h(4) = -16(16) + 128(4) \][/tex]
[tex]\[ h(4) = -256 + 512 \][/tex]
[tex]\[ h(4) = 256 \][/tex]
Thus, the maximum height is 256 feet.
### c. How long did it take for the rocket to reach the ground?
The rocket reaches the ground when the height [tex]\( h(t) = 0 \)[/tex]. We need to solve the equation:
[tex]\[ -16t^2 + 128t = 0 \][/tex]
Factor out the common term [tex]\( t \)[/tex]:
[tex]\[ t(-16t + 128) = 0 \][/tex]
This gives us two solutions:
[tex]\[ t = 0 \][/tex] (the time the rocket is launched) or
[tex]\[ -16t + 128 = 0 \][/tex]
[tex]\[ 16t = 128 \][/tex]
[tex]\[ t = \frac{128}{16} \][/tex]
[tex]\[ t = 8 \][/tex]
So, it takes 8 seconds for the rocket to return to the ground.
### Summary:
- The rocket reaches its maximum height after 4 seconds.
- The maximum height of the rocket is 256 feet.
- The rocket takes 8 seconds to reach the ground.
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