# Thread:

1. ## projectile motion

Let’s suppose you throw a ball straight up with an initial speed of 50 feet per second from a height of 6 feet.
1. Find the parametric equations that describe the motion of the ball as a function of time.
2. How long is the ball in the air?
3. Determine when the ball is at maximum height. Find its maximum height.

2. how far can you get?

3. I have all the answer correct until C. I cant figure out whether i need to factor, or simplify. I also know that the maximum height is the y coordinate of the vertex.

4. There are two ways to get your maximum height:

Method 1 (requires calculus)
i assume you ahve a parametric equation for the vertical motion like: y=f(t)

differentiate and find the value of t where $\frac{dy}{dt} = 0$. This will give you the time at which the ball is at its maximum height.
The height at that moment is then f(t).

Method 2
Consider the vertical motion of the ball only and use one of Newton's equations of motion:

$v=u + at$
v = 0 (at maximum height, it is no longer gaining altitude)
u = ? (should have been given in question)
a = accelleration (you should know the value of this!)
t = the time at which max height is reached (solve for this)

Then use another equation to get the height at this moment, eg
$v^2 = u^2 + 2as$

5. Originally Posted by kenzie103109
Let’s suppose you throw a ball straight up with an initial speed of 50 feet per second from a height of 6 feet.
1. Find the parametric equations that describe the motion of the ball as a function of time.
2. How long is the ball in the air?
3. Determine when the ball is at maximum height. Find its maximum height.
$y = 6 + 50t - 32t^2$

set $y = 0$ and solve for $t$ to determine how long the ball is airborne.

the graph of $y$ is parabolic ... use $t = \frac{-b}{2a}$ to find the vertex (time of max height)
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