The height h in metres above the ground of a projectile at any time t in seconds after the launch is defined by the function h(t)=-4t +48t +3
A.) Complete the square to write H in standard form
b.) Find the height of the projectile three seconds after the launch
c.) Find the maximum height reached by the projectile
D.) how many seconds after the launch is the maximum height reached
e.) what was the height of the projectile at the launch
f.) determine when the projectile hits the ground to the nearest tenth of a second
I only got a.) but not sure if its right
-4(t-6)^2+147
Question #2: Ashley was asked by her math teacher to find two numbers which differ by 8 and whose product is a minimum .
a.) if X represents the smaller number write a quadratic expression in x for the product of the two numbers
b.) write the product and completed square form
c.) determine the numbers and the minimum product
Thanks :)
55 is the product of Chau's height and
5
A.) To complete the square and write the function h(t) in standard form, follow these steps:
1. Start with the given function: h(t) = -4t + 48t + 3.
2. Group the t-terms together: h(t) = (-4t^2 + 48t) + 3.
3. Factor out the coefficient of t^2 from the first two terms: h(t) = -4(t^2 - 12t) + 3.
4. To complete the square, take half of the coefficient of t (-12) and square it: (-12/2)^2 = (-6)^2 = 36.
5. Add and subtract the calculated value inside the parenthesis: h(t) = -4(t^2 - 12t + 36 - 36) + 3.
6. Group and factor the terms inside the parenthesis: h(t) = -4((t - 6)^2 - 36) + 3.
7. Simplify: h(t) = -4(t - 6)^2 + 144 + 3.
8. Combine like terms: h(t) = -4(t - 6)^2 + 147.
Therefore, the function h(t) in standard form is -4(t - 6)^2 + 147.
B.) To find the height of the projectile three seconds after the launch, substitute t = 3 into the function h(t) and solve for h(3):
h(3) = -4(3 - 6)^2 + 147
= -4(-3)^2 + 147
= -4(9) + 147
= -36 + 147
= 111
Therefore, the height of the projectile three seconds after the launch is 111 metres.
C.) To find the maximum height reached by the projectile, observe that the maximum height occurs when the projectile is at the vertex of the parabolic path. The vertex can be found using the formula t = -b/2a, where a = -4 and b = 48 in the given function.
t = -b/2a
= -48/(2(-4))
= -48/(-8)
= 6
Therefore, the maximum height is reached at t = 6 seconds.
D.) The maximum height is reached at t = 6 seconds, as determined in the previous step.
E.) To find the height of the projectile at the launch (t = 0), substitute t = 0 into the function h(t) and solve for h(0):
h(0) = -4(0 - 6)^2 + 147
= -4(-6)^2 + 147
= -4(36) + 147
= -144 + 147
= 3
Therefore, the height of the projectile at the launch is 3 metres.
F.) The projectile hits the ground when the height, h(t), is equal to 0. To find this time, set h(t) = 0 and solve for t:
0 = -4(t - 6)^2 + 147
4(t - 6)^2 = 147
(t - 6)^2 = 147/4
t - 6 = ±√(147/4)
t - 6 = ±(√147/2)
t = 6 ± (√147/2)
Since we are looking for the positive value of t, we can ignore the negative solution.
t = 6 + (√147/2)
Evaluating this expression using a calculator, we find:
t ≈ 10.20 seconds
Therefore, the projectile hits the ground approximately 10.20 seconds after the launch.
A.) To complete the square and write the function in standard form, we can follow these steps:
1. Start with the given function: h(t) = -4t^2 + 48t + 3.
2. Divide the entire equation by -4 to make the coefficient of t^2 equal to 1: h(t) = t^2 -12t - (3/4).
3. Move the constant term (-3/4) to the right side of the equation: h(t) + (3/4) = t^2 - 12t.
4. Complete the square on the right side of the equation by adding the square of half the coefficient of t to both sides. The coefficient of t is -12, so the square of half of -12 is 36: h(t) + (3/4) + 36 = t^2 - 12t + 36.
5. Simplify the right side: h(t) + 147/4 = (t - 6)^2.
6. Rewrite the left side with a common denominator: (4h(t) + 147)/4 = (t - 6)^2.
7. Finally, write the equation in standard form: 4h(t) + 147 = 4(t - 6)^2.
So, the completed square form in standard form is: 4h(t) + 147 = 4(t - 6)^2.
B.) To find the height of the projectile three seconds after the launch (t = 3), substitute t = 3 into the function h(t):
h(3) = -4(3)^2 + 48(3) + 3
h(3) = -36 + 144 + 3
h(3) = 111
Therefore, the height of the projectile three seconds after the launch is 111 metres.
C.) To find the maximum height reached by the projectile, we need to find the vertex of the quadratic function. The vertex represents the highest point of the parabola. To find the vertex, we can use the formula: t = -b/2a.
In this case, a = -4 (coefficient of t^2) and b = 48 (coefficient of t). Calculating the vertex time t:
t = -48 / (2 * -4)
t = -48 / -8
t = 6
So, the maximum height is reached at t = 6 seconds.
D.) The maximum height is reached at t = 6 seconds.
E.) To find the height of the projectile at the launch, we substitute t = 0 into the function h(t):
h(0) = -4(0)^2 + 48(0) + 3
h(0) = 3
The height of the projectile at launch is 3 metres.
F.) The projectile hits the ground when the height h(t) is zero. So, to find when the projectile hits the ground, we set h(t) = 0 and solve for t:
0 = -4t^2 + 48t + 3
To find the approximate solutions, you can use a graphing calculator or the quadratic formula. Once you solve for t, you will get two values, and the earlier one will represent when the projectile hits the ground.