A rocket is launched from atop a 76-foot cliff with an initial velocity of 107 ft/s. The height of the rocket
above the ground at time t is given by .
A) When will the rocket hit the ground after it is launched? Round to the nearest tenth of a second.
(Solve Algebraically, Hint: Use the Quadratic Formula to solve and the round the square root answer one
decimal place)
B) When will the rocket reach 167ft from the ground?
The height equation that should have been given to you would be ...
height = -16t^2 + 107t + 76
"When will the rocket hit the ground". I will assume by ground you mean 76 below the cliff, so height = 0
-16t^2 + 107t + 76 = 0
16t^2 - 107t - 76 = 0
They told you to use the quadratic equation formula, so ...
t = (107 ± √(107^2 - 4(16)(-76)) )/32
= ..... you do the button pushing, use the positive result only
b)
-16t^2 + 107t + 76 = 167
16t^2 - 107t + 91 = 0
You will get 2 answers, both positive and both exact.
One will be the time it will reach a height of 167 on its way up, and the other
on its way down.
A. V = Vo + g*Tr
0 = 107 + (-32)Tr
Tr = 3.34 s. = Rise time.
V^2 = Vo^2 + 2g*h.
0 = 107^2 + (-64)h
h = 178.9 Ft. above roof.
ho+h = 76+178.9 = 254.9 Ft. above gnd.
0.5g*Tf^2 = 254.9
16Tf^2 = 254.9
Tf = 4 s. = Fall time.
Tr + Tf = 3.34 + 4 = 7.3 s. = Time to reach gnd.
B. 0.5g*T^2 = 254.9-167
16T^2 = 87.9
2.3 s.
To determine when the rocket will hit the ground after it is launched, we need to find the time at which the height of the rocket above the ground is equal to zero.
Given:
Initial height, h₀ = 76 ft
Initial velocity, v₀ = 107 ft/s
Height of the rocket above the ground at time t, h(t) = -16t² + v₀t + h₀
A) To find when the rocket hits the ground, we need to set h(t) = 0 and solve for t using the quadratic formula.
So, we have:
-16t² + 107t + 76 = 0
Using the quadratic formula: t = (-b ± √(b² - 4ac)) / 2a, where a = -16, b = 107, and c = 76.
t = (-107 ± √(107² - 4(-16)(76))) / (2(-16))
Simplifying further:
t = (-107 ± √(11449 + 4864)) / -32
t = (-107 ± √(16313)) / -32
Using a calculator, we find:
t ≈ (-107 ± 127.79) / -32
This gives us two potential solutions:
1) t ≈ (-107 + 127.79) / -32 ≈ 0.6652 s
2) t ≈ (-107 - 127.79) / -32 ≈ 7.8656 s
Since the rocket is launched at time t = 0, we can disregard the negative value of t. Therefore, the rocket will hit the ground approximately 0.6652 seconds after it is launched.
B) To find when the rocket reaches a height of 167 ft from the ground, we need to set h(t) = 167 and solve for t.
So, we have:
-16t² + 107t + 76 = 167
Simplifying further:
-16t² + 107t - 91 = 0
Using the quadratic formula, t = (-b ± √(b² - 4ac)) / 2a, where a = -16, b = 107, and c = -91.
t = (-107 ± √(107² - 4(-16)(-91))) / (2(-16))
Simplifying further:
t = (-107 ± √(11449 - 5824)) / -32
t = (-107 ± √(5625)) / -32
t = (-107 ± 75) / -32
Using the positive root:
t = (-107 + 75) / -32
Calculating the value:
t ≈ -32 / -32
t ≈ 1
The rocket will reach a height of 167 ft from the ground approximately 1 second after it is launched.
To find when the rocket will hit the ground, we need to determine the time when the height above the ground is zero.
The height of the rocket above the ground can be represented by a quadratic equation in the form: h(t) = -16t^2 + v0t + h0, where h(t) is the height above the ground at time t, v0 is the initial velocity of the rocket, and h0 is the initial height of the rocket.
Given that the initial velocity v0 is 107 ft/s and the initial height h0 is 76 ft, we get the equation: h(t) = -16t^2 + 107t + 76.
A) To find when the rocket hits the ground, we need to solve the equation h(t) = 0. We can do this by setting -16t^2 + 107t + 76 equal to zero and then solving for t. Using the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by x = (-b ± √(b^2 - 4ac))/2a, we have:
t = (-107 ± √(107^2 - 4(-16)(76)))/(2(-16))
After simplifying, we get:
t = (-107 ± √(11449 + 4864))/(−32)
t = (-107 ± √(16313))/(-32)
Calculating the square root of 16313, we get:
t = (-107 ± 127.78)/(-32)
Simplifying further, we have:
t = (-107 + 127.78)/(-32) and t = (-107 - 127.78)/(-32)
t = 20.78/(-32) and t = -234.78/(-32)
Thus, t ≈ -0.649 and t ≈ 7.336.
Since time cannot be negative in this context, the rocket will hit the ground approximately 7.3 seconds after it is launched (rounded to the nearest tenth of a second).
B) To find when the rocket reaches a height of 167 ft from the ground, we need to solve the equation h(t) = 167. We can do this by setting -16t^2 + 107t + 76 equal to 167 and then solving for t.
Setting -16t^2 + 107t + 76 = 167, we get:
-16t^2 + 107t + 76 - 167 = 0
-16t^2 + 107t - 91 = 0
We can now solve this quadratic equation by using the quadratic formula:
t = (-107 ± √(107^2 - 4(-16)(-91)))/(2(-16))
After simplifying, we have:
t = (-107 ± √(11449 - 5824))/(-32)
t = (-107 ± √(5625))/(-32)
Calculating the square root of 5625, we get:
t = (-107 ± 75)/(-32)
Simplifying further, we have:
t = (-107 + 75)/(-32) and t = (-107 - 75)/(-32)
t = -32/(-32) and t = -182/(-32)
Thus, t = 1 and t = 5.6875.
Therefore, the rocket will reach a height of 167 ft from the ground approximately 5.7 seconds after it is launched (rounded to the nearest tenth of a second).