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?

Transcript

00:04 All right, so we're given the equation for the height of a projectile, and we want to start by finding the velocity after two seconds and four seconds.

00:12 And so we're going to find the velocity in general, v of t, by finding the derivative of the height function.

00:18 So the derivative would be 24 .5 minus 9 .8t.

00:23 Now to find the velocity at 2, we can substitute 2 in there.

00:27 So we have 24 .5 minus 9 .8 times 2.

00:30 And that gives us 4 .9, and the units would be meters per second.

00:36 And the velocity at time 4 will just substitute a 4 in there.

00:42 And we get negative 14 .7 meters per second.

00:50 For part b, we want to find the time when the projectile reaches its maximum height.

00:55 And we have two ways to do this.

00:57 One way is sort of an algebra two way.

01:00 We recognize that this is a parabola, and it opens down, and we could find the vertex and the x -corps.

01:07 Coordinate of the vertex would be the time.

01:09 The other way is a calculus way, and we would realize that the velocity would be zero at the time when it's at its maximum point because we would have a horizontal tangent line there.

01:21 And so i say we do the calculus way since we're in calculus.

01:24 So we want the velocity to equal zero.

01:29 So we take our velocity 24 .5 minus 9 .8t and set it equal to zero, and we solve for t.

01:37 So we have 9 .8t equals 24.

01:39 4 .5 and we end up with t equals 2 .5 seconds.

01:47 So that is the time when the projectile is at the maximum height.

01:54 For part c we're finding the maximum height so we're going to take the time we just found which was 2 .5 seconds and substitute it into our height equation and we'll compute this and we end up with 32 .625 meters...

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