# Help!World Problem:Calculus III Projectile Motion (Velocity & Acceleration)

• Nov 28th 2007, 04:37 PM
googoogaga
Help!World Problem:Calculus III Projectile Motion (Velocity & Acceleration)

Assuming there is no resistance and knowing that a(t)= -9.8 m/s^2

Determine the maximum height and range of a projectile fired at a height of 1.5 meters above the ground with an initial velocity of a 100 meters per second and at an angle of 30 degrees above the horizontal.

maximum height: 129.1 meters
range: 886.3 meters
(how do you go about solving this problem using Vector-Valued Functions such as r(t)= x(t)i + y(t)j etc. )
• Nov 29th 2007, 02:44 AM
googoogaga
• Nov 29th 2007, 03:12 AM
topsquark
Quote:

Originally Posted by googoogaga

Assuming there is no resistance and knowing that a(t)= -9.8 m/s^2

Determine the maximum height and range of a projectile fired at a height of 1.5 meters above the ground with an initial velocity of a 100 meters per second and at an angle of 30 degrees above the horizontal.

maximum height: 129.1 meters
range: 886.3 meters
(how do you go about solving this problem using Vector-Valued Functions such as r(t)= x(t)i + y(t)j etc. )

The equation of motion can be expressed parametrically as you state with
$\displaystyle x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2$
and
$\displaystyle y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2$

(You can derive these from knowing that the acceleration components in both x and y directions are constant. Then $\displaystyle \frac{d^2x}{dt^2} = a_x$, etc.)

We need suitable initial conditions, of course. First off $\displaystyle a_x = 0$. Also the point $\displaystyle (x_0, y_0)$ acts as an origin for the problem. You can select any one that you like; for my purposes I'm going to choose $\displaystyle (x_0, y_0) = (0, 1.5~m)$. That way none of my y values are negative. We need to recognize the $\displaystyle v_{0x}$ and $\displaystyle v_{0y}$ as initial velocity components, so $\displaystyle v_{0x} = 100 \cdot cos(30)~m/s$ and $\displaystyle v_{0y} = 100 \cdot sin(30)~m/s$.

So the parametric equations of motion are:
$\displaystyle x = 100t~cos(30)$
and
$\displaystyle y = 1.5 + 100t~sin(30) - 4.9t^2$

We want the maximum height for the first problem. What is the condition for a moving object to be at an extremum? When the first derivative is 0, of course. So when $\displaystyle \frac{dy}{dt} = 0$ the particle is at its maximum height. As per my usual, I'll leave it to you to prove that this is indeed a maximum. (Remember you want y(t) not just t.)

The second problem asks about the range. This is the x value (actually $\displaystyle x - x_0$ value) for when the projectile is at ground level, that is to say y = 0 in my coordinate system. So you need to solve the system of equations:
$\displaystyle x = 100t~cos(30)$
and
$\displaystyle 0 = 1.5 + 100t~sin(30) - 4.9t^2$
for x.

-Dan