Question

Let’s suppose you throw a ball straight up with an initial speed of 50 feet per second from a height of 6 feet.

Find the parametric equations that describe the motion of the ball as a function of time.
How long is the ball in the air?
Determine when the ball is at maximum height. Find its maximum height.

Answer 1

So start with a basic kinematic equation:
\[d = v_{i}*t + \frac{1}{2}*g*t^{2} \]
Since the ball is thrown straight up , it is only moving in the y direction, so we can write the parametric equations as follows:
\[y(t) = 50*t - \frac{1}{2}*32.17*t^2 + 6\]\[x(t) = 0\]
Where g = -32.17 ft/s^2. So to find how long the ball was in the air, we solve y(t) = 0 (when it's on the ground d=0):
\[0 = - \frac{1}{2}*32.17*t^2 + 50*t + 6\]
\[t = -0.116, 3.224\]
So the ball stays in the air for 3.22 seconds. To find it's maximum height, we must find where the derivative is 0:
\[y'(t) = -32.17*t + 50\]
\[t = \frac{-50}{-32.17} \approx 1.554\]
So the ball reaches its maximum height at t=1.55 s. To find what the maximum height is we plug in 1.55 into the original equation:
\[y(1.55) = 50*(1.55) - \frac{1}{2}*32.17*(1.55)^2 + 6 \approx 44.856\]
So the ball reaches a distance of 44.86 ft.

Answer 2

i got y=-16t^2 + 50t + 6