Problem

If a ball is thrown vertically upward with a velo…

03:24

Problem 7 Hard Difficulty

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?

Answer

a) $-14.7 \mathrm{m} / \mathrm{s}$
b) $2.5 \mathrm{s}$
c) 32.625 $\mathrm{m}$
d) $32.625 \mathrm{m}\left[\text { or } 32 \frac{5}{8} \mathrm{m}\right]$
e) $v\left(t_{f}\right)=24.5-9.8 t_{f} \approx-25.3 \mathrm{m} / \mathrm{s}[\text { downward }]$

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Video Transcript

All right, So we're given the equation for the height ever projectile, and we want to start by finding the velocity after two seconds and four seconds. And so we're going to find the velocity in General V of tea by finding the derivative of the height function. So the derivative would be 24.5 minus 9.8 t Now, to find the velocity it to weaken substitute to in there. So we have 24.5 minus 9.8 times two, and that gives us 4.9 and the units would be meters per second and the velocity at time four will just substitute a four in there and we get negative 14.7 meters per second for part B. We want to find the time when the projectile reaches its maximum height and we have two ways to do this. One way is sort of an algebra two way. We recognize that this is a parabola and it opens down and we could find the Vertex and the X coordinate of the Vertex would be the time. 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. And so I say we do the calculus way since we're in calculus. So we want the velocity equals zero. So we take our velocity 24.5 minus 9.80 and said it equal to zero. And we saw for teeth. So we have 9.8 t equals 24.5 and we end up with t equals 2.5 seconds. So that is the time when the projectile is at the maximum height 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 substituted into our height equation and will compute this and we end up with 32.6 to 5 meters for part D. We want to find the time when it hits the ground. So when it hits the ground, its height is zero. So we're going back to the height equation on. We're setting it equal to zero. I'm going to rearrange the terms of little bits. We have negative 4.9 t squared plus 24.5 t plus two equals zero. Now, unfortunately, this is not factory ble. So we're going to solve this equation with the quadratic formula, and I'm going to start by multiplying the whole equation by negative one just so that we can lead with a positive coefficient. Just seems to make life a little bit easier. Now, we're going to use the quadratic formula so we have t equals the opposite of B over two, a plus or minus the square root of B squared minus four a. C all over two a. And you're gonna simplify that and work that out and we end up with 2.5 plus or minus the square root of 6 39.45 over 9.8 and you're gonna put that in your calculator and you get to answers approximately 5.8 or negative 0.8 But you realize that you're not going to keep the negative time in this context. So just the positive one now, because for party, we're going to use this number for something else. We might store that in our calculators, so that we will be more accurate for part e for party were finding the velocity when it hits the ground and we use our velocity equation and we substitute that number we just got in the previous part the approximate values 5.0, wait. It would be a good idea to use the stored value so you don't lose too much accuracy. So we have 24.5, minus 9.8 times five point a 5.0.8 and we end up with approximately negative 25.3 meters per second.

Heather Z.

Oregon State University

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Problem

If a ball is thrown vertically upward with a velo…

Problem 7 Hard Difficulty

Answer

a) $-14.7 \mathrm{m} / \mathrm{s}$
b) $2.5 \mathrm{s}$
c) 32.625 $\mathrm{m}$
d) $32.625 \mathrm{m}\left[\text { or } 32 \frac{5}{8} \mathrm{m}\right]$
e) $v\left(t_{f}\right)=24.5-9.8 t_{f} \approx-25.3 \mathrm{m} / \mathrm{s}[\text { downward }]$

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Calculus: Early Transcendentals

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Calculus: Early Transcendentals

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Precalculus Review - Intro

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a) $-14.7 \mathrm{m} / \mathrm{s}$b) $2.5 \mathrm{s}$c) 32.625 $\mathrm{m}$d) $32.625 \mathrm{m}\left[\text { or } 32 \frac{5}{8} \mathrm{m}\right]$e) $v\left(t_{f}\right)=24.5-9.8 t_{f} \approx-25.3 \mathrm{m} / \mathrm{s}[\text { downward }]$

Calculus: Early Transcendentals

Kayleah T.

Caleb E.

Samuel H.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

Kayleah T.

Caleb E.

Samuel H.

Joseph L.

a) $-14.7 \mathrm{m} / \mathrm{s}$
b) $2.5 \mathrm{s}$
c) 32.625 $\mathrm{m}$
d) $32.625 \mathrm{m}\left[\text { or } 32 \frac{5}{8} \mathrm{m}\right]$
e) $v\left(t_{f}\right)=24.5-9.8 t_{f} \approx-25.3 \mathrm{m} / \mathrm{s}[\text { downward }]$