Hello, I am having a lot of trouble being certain as to if I'm right or not in my solution.
Determine the range, arc length, and maximum height of a projectile fired from a height (h) of 4 feet above ground, initial velocity (v) of 80 ft/sec at an angle (x) of 30 degrees with the horizontal. Use the following formulas for parametric equations.
x = (v * cos(x))t
y = h + (v * sin(x))t - 16t^2
Using a calculator to graph both of these equations yields an upside down parabola, of course. I'm just not sure if my procedure is correct in solving the question.
I just substituted the given values from the start.
x = 40 * sqrt(3) * t
y = -16t^2 + 40t + 4
I figured that the projectile would reach the ground when y(t) is equal to zero. I thought about eliminating the parameter, which would make the problem easier for me to understand, but I don't think my instructor wants us to do that. So I set the y equation equal to zero.
-16t^2 +40t + 4 = 0
t = -.05 and 2.55
These two values for the parameter get me to the first x-intercept and the maximum of the parabola, but I'm not sure why it worked that way, I thought it would get me to both of the x-intercepts.
So reasoning this way, I plugged the t = 2.55 into the y(t) to find the maximum height, which turned out to be 105.235 feet.
Since I don't know what to do to get the range (the starting value to the value that it reaches the ground), I doubled the parameter to get 5.1, which my graph showed me it to hit the ground at that time, and plugging 5.1 in for x(t), I got a distance of 353.338 feet.
And the arc length, I don't have much trouble with, however, I'm not sure if it's right because I'm not sure if my range is correct.
Arc Length = integral of sqrt [ (40 * sqrt(3))^2 + (-32t + 40)^2 ] dt, with a lower limit of 0 (the time it was fired) and an upper limit of 5.1 (the time I think it hit the ground).
I just feel like I was guessing through this entire problem without using an exact method. If this isn't correct, can somebody push me into a better direction?
Thank You!!!!
Determine the range, arc length, and maximum height of a projectile fired from a height (h) of 4 feet above ground, initial velocity (v) of 80 ft/sec at an angle (x) of 30 degrees with the horizontal. Use the following formulas for parametric equations.
x = (v * cos(x))t
y = h + (v * sin(x))t - 16t^2
Using a calculator to graph both of these equations yields an upside down parabola, of course. I'm just not sure if my procedure is correct in solving the question.
I just substituted the given values from the start.
x = 40 * sqrt(3) * t
y = -16t^2 + 40t + 4
I figured that the projectile would reach the ground when y(t) is equal to zero. I thought about eliminating the parameter, which would make the problem easier for me to understand, but I don't think my instructor wants us to do that. So I set the y equation equal to zero.
-16t^2 +40t + 4 = 0
t = -.05 and 2.55
These two values for the parameter get me to the first x-intercept and the maximum of the parabola, but I'm not sure why it worked that way, I thought it would get me to both of the x-intercepts.
So reasoning this way, I plugged the t = 2.55 into the y(t) to find the maximum height, which turned out to be 105.235 feet.
Since I don't know what to do to get the range (the starting value to the value that it reaches the ground), I doubled the parameter to get 5.1, which my graph showed me it to hit the ground at that time, and plugging 5.1 in for x(t), I got a distance of 353.338 feet.
And the arc length, I don't have much trouble with, however, I'm not sure if it's right because I'm not sure if my range is correct.
Arc Length = integral of sqrt [ (40 * sqrt(3))^2 + (-32t + 40)^2 ] dt, with a lower limit of 0 (the time it was fired) and an upper limit of 5.1 (the time I think it hit the ground).
I just feel like I was guessing through this entire problem without using an exact method. If this isn't correct, can somebody push me into a better direction?
Thank You!!!!