The question I have is to solve the problem.
A model rocket is launched from the ground with an initial speed of 50 feet per second. The equation that models its height, h feet, off the ground t seconds after it was fired is
h=-16ft^2+50t
a)How high is the rocket 1 seconds after it was fired?
b)How long will it take the rocket to return to earth?
a. h = 50*1 - 16*1^2 = 34 Ft.
b. V = Vo + g*t
Tr = (V-Vo)/g = (0-50)/-32 = 1.56 s. =
Rise time.
Tf = Tr = 1.56 s. = Fall time.
T = Tr + Tf = 1.56 + 1.56 = 3.125 s. =
Time t0 return to earth.
t
To solve this problem, we will use the given equation h = -16t^2 + 50t.
a) To find how high the rocket is 1 second after it was fired, we substitute t = 1 into the equation:
h = -16(1)^2 + 50(1)
h = -16 + 50
h = 34 feet
Therefore, the rocket is 34 feet high 1 second after it was fired.
b) To find how long it takes for the rocket to return to the ground, we need to find the value of t when h = 0.
0 = -16t^2 + 50t
To solve this quadratic equation, we can factor it or use the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = -16, b = 50, and c = 0. Plugging in these values:
t = (-50 ± √(50^2 - 4(-16)(0))) / (2(-16))
t = (-50 ± √(2500)) / (-32)
t = (-50 ± 50) / (-32)
We have two possible solutions:
1) t = (-50 + 50) / (-32) = 0 / (-32) = 0
2) t = (-50 - 50) / (-32) = -100 / (-32) = 3.125
Therefore, it will take approximately 3.125 seconds for the rocket to return to the ground.
Note: In this case, we discard the negative solution because time cannot be negative in this context.
To solve these questions, we will use the given equation for the height of the rocket, h = -16t^2 + 50t, where h is the height in feet and t is the time in seconds.
a) To find how high the rocket is 1 second after it was fired, we substitute t = 1 into the equation h = -16t^2 + 50t:
h = -16(1)^2 + 50(1)
h = -16 + 50
h = 34 feet
Therefore, the rocket is 34 feet high 1 second after it was fired.
b) To find out how long it will take for the rocket to return to earth, we need to find the value of t for which the height, h, is equal to zero. This indicates that the rocket has returned to the ground.
We set h = -16t^2 + 50t = 0 and solve for t:
-16t^2 + 50t = 0
t(-16t + 50) = 0
Setting each term equal to zero:
t = 0 (which represents the time at the start when the rocket was launched)
-16t + 50 = 0
-16t = -50
t = 50/16
t ≈ 3.125
Therefore, it will take approximately 3.125 seconds for the rocket to return to the earth.