The function H(t) = -16t2 + vt + s shows the height H (t), in feet, of a projectile launched vertically from s feet above the ground after t seconds. The initial speed of the projectile is v feet per second.
Part A: The projectile was launched from a height of 90 feet with an initial velocity of 50 feet per second. Create an equation to find the time taken by the projectile to fall on the ground. (2 points)
Part B: What is the maximum height that the projectile will reach? Show your work. (2 points)
Part C: Another object moves in the air along the path of g(t) = 28 + 48.8t where g(t) is the height, in feet, of the object from the ground at time t seconds.
Use a table to find the approximate solution to the equation H(t) = g(t), and explain what the solution represents in the context of the problem? [Use the function H(t) obtained in Part A, and estimate using integer values] (4 points)
Part D: Do H(t) and g(t) intersect when the projectile is going up or down, and how do you know?
Part A: To find the time taken by the projectile to fall on the ground, we need to set H(t) equal to zero and solve for t. In this case, the initial height, s, is 90 feet, and the equation becomes:
-16t^2 + vt + 90 = 0
Part B: To find the maximum height reached by the projectile, we need to find the vertex of the parabolic function represented by H(t). The vertex can be found using the equation:
t = -b / (2a)
In this case, a is -16 and b is v (which is 50 feet per second). Plugging in these values, we get:
t = -50 / (2 * -16)
t = 50 / 32
t ≈ 1.5625 seconds
To find the maximum height, we substitute this value of t back into the equation H(t):
H(1.5625) = -16(1.5625)^2 + v(1.5625) + s
Part C: To find the approximate solution to the equation H(t) = g(t), we need to create a table of integer values and evaluate the heights for each value of t.
Let's create a table using integer values for t, and calculate the corresponding heights for both H(t) and g(t):
t | H(t) | g(t)
--------------
0 | 90 | 28
1 | 54 | 76.8
2 | 18 | 124.6
3 | -18 | 172.4
4 | -54 | 220.2
From the table, we can see that the approximate solution to the equation H(t) = g(t) is t ≈ 2 seconds. This represents the time at which both objects (the projectile and the other object moving along g(t)) have the same height above the ground.
Part D: To determine if H(t) and g(t) intersect when the projectile is going up or down, we need to analyze the signs of their respective velocities.
For H(t), the coefficient of t is -16, which means the velocity is negative. This indicates that the projectile is moving downward when t > 0.
For g(t), the coefficient of t is 48.8, which means the velocity is positive. This indicates that the other object moving along g(t) is moving upwards.
Therefore, H(t) and g(t) intersect when the projectile is going down because H(t) represents the height of the projectile, and g(t) represents the height of another object moving upwards.
Part A: To find the time taken by the projectile to fall on the ground, we need to set H(t) equal to zero since the height of the projectile at the ground is zero. So, we can write the equation as:
-16t^2 + vt + s = 0
Substituting the given values, we have:
-16t^2 + 50t + 90 = 0
This is the equation to find the time taken by the projectile to fall on the ground.
Part B: To find the maximum height that the projectile will reach, we need to determine the vertex of the parabolic function H(t) = -16t^2 + vt + s. The vertex of a parabola is given by (-b/2a, H(-b/2a)). In this case, a = -16, b = v, and s = 90. Substituting these values, we have:
t = -v / (2 * -16) = -v / -32 = v / 32
H(v/32) = -16(v/32)^2 + v(v/32) + 90
Simplifying this equation will give us the maximum height that the projectile will reach.
Part C: To find the approximate solution to the equation H(t) = g(t), we can create a table by substituting different integer values for t and evaluating the corresponding values of H(t) and g(t).
For example:
t | H(t) | g(t)
---------------
1 | |
2 | |
3 | |
4 | |
5 | |
Substitute each value of t into the equations H(t) = -16t^2 + vt + s and g(t) = 28 + 48.8t to find the corresponding values of H(t) and g(t). Once we find the common value for H(t) and g(t), that will be the approximate solution to the equation. The solution represents the time at which both objects will be at the same height.
Part D: To determine if H(t) and g(t) intersect when the projectile is going up or down, we need to compare the slopes of the functions. The slope of H(t) is given by the derivative of -16t^2 + vt + s with respect to t, while the slope of g(t) is given by the derivative of 28 + 48.8t with respect to t.
If the slopes have the same sign (both positive or both negative), then H(t) and g(t) intersect while the projectile is either going up or down. If the slopes have opposite signs (one positive and one negative), then H(t) and g(t) intersect while the projectile is at its highest point (vertex).
A. V = Vo + g*Tr = 0, Tr = -Vo/g = Rise time.
T1 = Tr + Tf = Time to rise from launching point(90Ft.) to max ht. and return to launching point. Tr = Tf; Therefore, T1 = 2Tr = -2Vo/g.
Vo*T2 - 16T2^2 = 90, 50T2-16*50^2 = 90, 50T2-40,000 = 90,
50T2 = 40,090, T2 = 802 Seconds To fall 90 Ft. to gnd.
T = T1+T2, T = -2Vo/g + 802 = Time to rise to max ht. and fall to gnd.
T = (-100/-32) + 802 = 805 Seconds.
B. V^2 = Vo^2 + 2g*h = 0, h = -Vo^2/2g, Vo = 50 Ft/s, g = -32 Ft/s^2, h = ?.