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4. One kicks a football upward. On the way up, it passes the roof of a building at height $h$ in time $t$ after the kick. What is the total time the football is in the air, from when it was kicked to when it lands back on the ground? (1) $\frac{2h}{gt} + t$ (2) $\frac{2h}{gt} - t$ (3) $\frac{h}{gt} + t$ (4) $\frac{h}{gt} - t$ (5) $\frac{h}{gt} + 2t$ Make y-axis go up and y=0 at the ground level. 1D motion with constant acceleration: $y_f = y_i + v_it + \frac{1}{2}at^2$ In this problem: At the time the ball passes the roof: $h = 0 + v_it - \frac{1}{2}gt^2$ At the time when the ball lands back on ground: $0 = 0 + v_iT - \frac{1}{2}gT^2$ Two equations, two unknowns ($v_i$ and $T$): one can solve for both.

          4. One kicks a football upward. On the way up, it passes the roof of a building at height $h$ in time $t$ after the kick. What is the total time the football is in the air, from when it was kicked to when it lands back on the ground?
(1) $\frac{2h}{gt} + t$
(2) $\frac{2h}{gt} - t$
(3) $\frac{h}{gt} + t$
(4) $\frac{h}{gt} - t$
(5) $\frac{h}{gt} + 2t$
Make y-axis go up and y=0 at the ground level.
1D motion with constant acceleration:
$y_f = y_i + v_it + \frac{1}{2}at^2$
In this problem:
At the time the ball passes the roof:
$h = 0 + v_it - \frac{1}{2}gt^2$
At the time when the ball lands back on ground:
$0 = 0 + v_iT - \frac{1}{2}gT^2$
Two equations, two unknowns ($v_i$ and $T$): one can solve for both.
        
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4. One kicks a football upward. On the way up, it passes the roof of a building at height h in time t after the kick. What is the total time the football is in the air, from when it was kicked to when it lands back on the ground?
(1) (2h)/(gt) + t
(2) (2h)/(gt) - t
(3) (h)/(gt) + t
(4) (h)/(gt) - t
(5) (h)/(gt) + 2t
Make y-axis go up and y=0 at the ground level.
1D motion with constant acceleration:
yf = yi + vit + (1)/(2)at^2
In this problem:
At the time the ball passes the roof:
h = 0 + vit - (1)/(2)gt^2
At the time when the ball lands back on ground:
0 = 0 + viT - (1)/(2)gT^2
Two equations, two unknowns (vi and T): one can solve for both.

Added by Loretta C.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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One kicks a football upward. On the way up, it passes the roof of a building at height h in time t after the kick. What is the total time the football is in the air, from when it was kicked to when it lands back on the ground? 2h (1) + t It 2h (2) gt h. 3 h h (5) + 2t gt 4 26 Make y-axis go up and y=0 at the ground level 1D motion with constant acceleration: Yf = Yi + vit + at In this problem: h = 0 + vt - 0.5gt^2 At the time the ball passes the roof: At the time when the ball lands back on the ground Two equations, two unknowns (v and t): one can solve for both.
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Transcript

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00:01 It looks like in this question that the equations are given to us.
00:06 And so in part a, as i'm trying to write here, in part a, we've got a football player with an angle of 36 degrees.
00:20 So alpha is 36 degrees and a v -sub -0 of 60 feet per second.
00:49 And then we're going to use g is 32 .2, no, 32 feet per second.
00:59 All right.
01:00 So, x is going to be v0 cosine alpha, 60 cosine of 36, v0 cosine alpha, 20 cosine of 36, v0 cosine alpha times t.
01:25 Y is going to be v0 sine alpha t minus one -half times 32 times t squared.
01:52 Graph the path that the football follows.
01:56 Well, i know that a football will follow a parabola.
02:03 I already know this from physics class.
02:08 So, x, y, it would be nice to know the maximum point.
02:27 So, i'm going to do dy over dt equals zero.
02:42 So that's just going to be 60 sine of 36 minus half.
02:53 Of 32 is 16 but then i have to multiply by 2 again so it's going to be 32 t and for the maximum i want that to be zero and so the maximum point is going to be at t equals 60 sine 36 over 32 just rearranging that for t i let me put that in a calculator to see exactly what time that is.
03:33 Now i need to make sure i set my calculator for degrees.
03:41 There we go.
03:47 60 sign of 36 divided by 32.
03:55 So that's going to be at 1 .10 to 1 .1 to 1 second.
04:08 The x value at that time is just going to be 60 cosine of 36 times that time.
04:19 So i'm going to take that time, and i'm going to multiply that times 60 cosine 36.
04:26 I'm going to multiply that times 1 .1021.
04:37 Okay, and that time, no, that x value is going to be 53 .5 feet.
04:58 The y value at that time is going to be 60, sine 36t minus 16 t squared.
05:22 So let's figure that out.
05:28 60, sign, 36 times time minus 16 times time, times time, squared.
05:50 19 .4.
06:02 All right.
06:03 So now we can graph it pretty easily.
06:06 This is going to be, let's say that this is 50.
06:10 This would be 100 out here.
06:15 It goes over 53, 54, and up 20.
06:22 Like that.
06:27 So the height here is about just under 20, 20 feet.
06:40 It's also in feet over here.
06:44 And it's going to land at 53 .5 times 2, which is going to be 107.
06:50 It's going to land at 107 feet...
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