anonymous
  • anonymous
A projectile is launched upward from the ground at 60 m/s. a. How long will it take to reach its highest point? b.) How high does it go? c.) How long does it take to hit the ground?
Calculus1
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Time = u sin 60 /g H = u^2 sin^2 60 /2g T = 2u sin 60/g
anonymous
  • anonymous
Thank you but how did you get time and what does the u stand for?
anonymous
  • anonymous
I'm sorry I don't understand the u?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
u = initial Velocity
anonymous
  • anonymous
Thank you but I'm still confused.
anonymous
  • anonymous
Okay, we start out with the fundamental idea that everything falls at constant acceleration. We'll call this constant acceleration (with respect to time) \(a\). We'll plug it in for the real value later.
anonymous
  • anonymous
The velocity is the going to be the anti derivative: \[ v(t) = \int a dt = at+C \]We need to find the constant of integration. Suppose the initial velocity is \(v_0\) (we'll plug it in later). \[ v(0) = v_0 = a(t) +C = C \]So our constant of integration is just the initial velocity \(v_0\). We now have the following equation: \[ v(t) = at + v_0 \]Which is enough to answer question part a
anonymous
  • anonymous
a. How long will it take to reach its highest point? <-- Well, the highest point is going to be the maximum position. Remember to find the maximum of a function, you need to find the critical number (when the derivative equals 0). Since velocity is the derivative of position with respect to time, we just need to set it to 0 to find critical numbers: \[ 0 = at + v_0 \implies at = -v_0 \implies t = -v_0/a \]So the amount of time it will take is just the initial velocity divided by acceleration. Don't worry about that negative sign. Since initial velocity is upward and gravity is downward, the negatives will cancel out. We will get a positive time value.
anonymous
  • anonymous
b.) How high does it go? <-- For this we need our position function. Again we find the anti-derivative.\[ x(t) = \int v(t)dt = \int at+v_0dt = \frac{1}{2}at^2+v_0t+C \]We have another constant of integration. Again we'll suppose the initial distance of \(x_0\). \[ x(0) = x_0 = \frac{1}2{}a(0)^2 + v_0(0) + C = C \]So once again, our constant of integration is just the initial value. \[ x(t) = \frac{1}{2}at^2+v_0t+x_0 \] We know it will reach it's highest value at \(t = -v_0/a\), so let's plug that in: \[ \begin{split} x(-v_0/a) &= \frac{1}{2}a(-v_0/a)^2+v_0(-v_0/a)+x_0 \\ &= \frac{1}{2}\frac{v_0^2}{a}-\frac{v_0^2}{a}+x_0 \\ &= -\frac{1}{2}\frac{v_0^2}{a}+x_0 \\ &= -\frac{v_0^2}{2a}+x_0 \\ \end{split} \] So we have our max distance being: \[ x_{max} = -\frac{v_0^2}{2a}+x_0 \]
anonymous
  • anonymous
Okay, I understand that.
anonymous
  • anonymous
Now how do I apply real numbers?
anonymous
  • anonymous
Like where does the sixty actually belong?
anonymous
  • anonymous
c.) How long does it take to hit the ground? <-- So we will call the position of the ground \(x_g\). We need to solve for \(t\) given \(x(t) = x_g\)\[ x_g = \frac{1}{2}at^2 + v_0t +x_0 \implies 0 = \frac{1}{2}at^2 + v_0t +x_0 -x_g \]This is just the root of a quadratic equation. \[ \begin{split} t &= \frac{-(v_0)\pm \sqrt{(v_0)^2-4(\frac{1}{2}a))(x_0-x_g)}}{2(\frac{1}{2}a)} \\ &= \frac{-v_0\pm \sqrt{v_0^2-2a(x_0-x_g)}}{a} \end{split} \] This is giving us two times. The earlier time is just when the projectile left the ground, we want the later time, when it hits the ground. Since the \(\sqrt{\ }\) is going to yield a positive value, we set the \(\pm \) to \(+\) to get the later time. \[ t=\frac{-v_0 + \sqrt{v_0^2-2a(x_0-x_g)}}{a} \]
anonymous
  • anonymous
Now we can plug in actual numbers. What we need are \[ a \\ v_0 \\ x_0 \\ x_g \]
anonymous
  • anonymous
Let's say that upward means 'positive' position, and downward means 'negative' position. \(a\) is gravitational acceleration downward. \(-9.8m/s^2\) \(v_0\) is our initial velocity: \(60m/s\) \(x_0\) is our initial position: \(0m\) \(x_g\) is the position of the ground: \(0m\)
anonymous
  • anonymous
Taking a second look... I think we should try both signs of \(\pm\) rather than just the \(+\), my assumption that that \(+\) version would be higher was incorrect.
anonymous
  • anonymous
Recap: \[ \begin{array}{rcl} t_{top} &=& -v_0/a \\ x_{top} &=& -\frac{v_0^2}{2a}+x_0 \\ t_{ground} &=& \max\left(\frac{-v_0+ \sqrt{v_0^2-2a(x_0-x_g)}}{a}, \frac{-v_0- \sqrt{v_0^2-2a(x_0-x_g)}}{a} \right) \\ a &=& -9.8m/s^2 \\ v_0 &=& 60m/s \\ x_0 &=& 0m \\ x_g &=& 0m \end{array} \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.