Maximum height?

Question

If a baseball is projected upward from ground level with an initial velocity of 64 feet per second, then its height is a function of time, given by s(t) = -16t^2 + 64t. Graph this fuction for 0 is less than or equal to t is less than or equal to 4. What is the maximum height reached by the ball? Please explain your answer. Thanks!

Answer 1

If you differntiate the equation you get;

s(t)dt = -32t + 64,

when this equation = 0 it gives you the time that it takes the ball to hit it's maximum height.

-32t + 64 = 0 => 32t = 64 => t = 2.

Plug this into the original equation to get the height;

s(t) = -16*2^2 + 64*2 => -64 + 128 = 64 feet.

i hope that helps.

Answer 2

In this kind of problem you have 3 things to look at.
Position, velocity, and accelleration.

s(t) = 16t^2 + 63t + 0

Look at your equation as these three things interacting.
16t^2 is the distance as a function of time due to accelleration.
64t is the distance as a function of time of the initial velocity.
and there is the 0 at the end I added, which is the initial position ... or displacement(s)

If you graph this function you will see the ball rise up ... slowing down until at the top of its arc it will stop due to the accelleration component, and then start to fall down again as the accelleration of gravity overcomes the initial projected velocity.

I'm not going to tell you the answer, but you can do this several ways. Look for the maximum of S(t) ... differentiate and set to 0.

This is also looking at where the velocity is 0 ... right at the top of the arc ... or you can graph it and try to read the graph and tell at what time the ball is at its maximum height.

Answer 3

Throwing it straight up is like dropping it straight down (or throwing it horizontally):

Since velocity = acceleration * time, time = 64/32 = 2 seconds.

Then distance = 1/2 * a * t^2 = 16 * 4 = 64 feet.