The height (in meters) of a projectile shot vertically...

The height (in meters) of a projectile shot vertically upward from a point $2 \mathrm{~m}$ above ground level with an initial velocity of $24.5 \mathrm{~m} / \mathrm{s}$ is $h=2+24.5 t-4.9 t^{2}$ after $t$ seconds. (a) Find the velocity after $2 \mathrm{~s}$ and after $4 \mathrm{~s}$. (b) When does the projectile reach its maximum height? (c) What is the maximum height? (d) When does it hit the ground? (e) With what velocity does it hit the ground?

The height (in meters) of a projectile shot vertically upward from a point $2 \mathrm{~m}$ above ground level with an initial velocity of $24.5 \mathrm{~m} / \mathrm{s}$ is $h=2+24.5 t-4.9 t^{2}$ after $t$ seconds.
(a) Find the velocity after $2 \mathrm{~s}$ and after $4 \mathrm{~s}$.
(b) When does the projectile reach its maximum height?
(c) What is the maximum height?
(d) When does it hit the ground?
(e) With what velocity does it hit the ground?

Key Concepts

Finding Maximum Height

Determining the maximum height of a projectile involves finding the time at which the vertical velocity becomes zero. This is because, at the peak of its trajectory, the projectile momentarily stops moving upward before beginning to descend, marking a critical point in its motion.

Solving Quadratic Equations

The height function in projectile problems is often quadratic in time. Solving these quadratic equations, particularly by setting the height equal to a given value (such as ground level), allows one to determine key time points in the projectile's trajectory, like the time of impact.

Kinematics Equations

Kinematics involves the equations that relate displacement, velocity, acceleration, and time. In the context of projectile motion, these equations help model the behavior of an object under constant acceleration due to gravity, allowing for predictions of position and velocity at any given time.

Projectile Motion

Projectile motion refers to the movement of an object that is projected into the air and is influenced only by gravity (ignoring air resistance). This creates a parabolic trajectory, and understanding this concept helps in analyzing the motion in both the vertical and horizontal directions, even when only vertical motion is considered.

Differentiation for Velocity

Differentiation is a mathematical tool used to compute the rate of change of a function. In projectile motion, differentiating the height (position) function with respect to time yields the velocity function, which is crucial for determining the speed and direction of the projectile at specific instants.

Transcript

00:01 Okay, so for this problem, we are given that a projectile shot vertically upward from a 0 .2 meters above the ground with an initial velocity of 24 .5 meters per second is given by our equation seen here.

00:17 So that is the position equation that we're given.

00:20 So we are asked to find what is the velocity at two seconds and what is the velocity at four seconds? so keep in mind.

00:29 Let's make a little note.

00:30 Over here that h of t, then h prime of t is going to be equal to our velocity.

00:42 Then for kicks and giggles, we'll go ahead and do h double prime of t.

00:45 We won't meet that in this case, but we, oh, and this should not be, v prime.

00:54 So it goes h of t, h prime of t is equivalent to our velocity, and h double prime of t is equivalent to our acceleration.

01:02 And so if we're asked to find the velocity at 2 and at 4, we want to find what is h prime of t.

01:11 So let's just go ahead and do that real fast.

01:14 So h prime of t is going to be equal to 24 .5 minus 9 .8t.

01:24 And so now that we have our v of 2, v of 4, this is also equal to v of t.

01:32 We're going to put 24 .5 times minus 9 .8 times 2 is going to be equal to our b of.

01:47 And then 24 .5 minus 9 .8 times 4.

01:54 It's going to be equal to our v of 4.

01:59 Once we evaluate those out, we get that v of 2 is equal to 4.

02:05 Point nine and v of four is equal to negative 14 .7.

02:10 So at two, our velocity is positive, which means the ball is going in an upwards direction.

02:17 And at t equals four, we are in a downwards trajectory towards the ground.

02:24 So that is going to be the answer for part a.

02:28 So now we're asked, when does it reach its maximum height? so when we're thinking about maximums, we're thinking about where are our critical values first and foremost? and are these critical values, maximums, or minimums? so what's going to tell us this is going to be our h prime of t when that is equal to zero.

02:49 So we're going to have, whoops, 24 .5 minus 9 .8t is going to be equal to zero.

03:00 And we do want to find where this t is.

03:03 So after doing some calculations, we get, t equals 2 .5.

03:09 If we put that in a number line just to check, we have zero on the left, three on the right.

03:15 If we plug in zero into this, we get that that's a positive.

03:19 We plug three into the same equation...

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