Step 4 25.5 Substituting the values we know into the equation for the y-component of displacement above, gives us the following result. Remember that ay = -g. Be careful with signs! 0+ 18.7 ✔m/s) (2.6 s) + ²/([ ✔ m/s²)([ X y = Your response differs from the correct answer by more than 10%. Double check your calculations. s²) -9.81 18.7 m/s. 36 x Your response differs from the correct answer by more than 100%. m

Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter3: Motion In Two Dimensions
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A place-kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball
leaves the ground with a speed of 23.3 m/s at an angle of 53.5° to the horizontal.
(a) By how much does the ball clear or fall short of clearing the crossbar? (Enter a negative answer if it falls short.)
(b) Does the ball approach the crossbar while still rising or while falling?
Step 1
(a) The football is moving with constant velocity in the x-direction. We denote the initial horizontal position of the football by xo and its initial x-component of velocity by Vox. For the position
in the horizontal direction in terms of time t, we have
x = xo + Voxt.
Solving for t gives the following equation.
t =
We do not know the horizontal component of velocity but we can calculate it from the angle of the football's path to the horizontal. Remember that we use trigonometry to find horizontal and
vertical components of a vector when we know its magnitude at a particular angle. We have
23.3 m/s cos 53.5 ✔✔✔
1/s) co
53.5 13.86✔
Vox
x-xo
Vox
t =
=
23.3✔
Step 2
We are looking for the time for the ball to reach the goal, wherex - X is the distance to the goal. Substituting the value we found for Vox into the equation for t above, gives the following for
t.
36
13.9
m) - (@ 0✔
36 m
13.9 m/s
0 m
= 2.59✔
13.9 m/s.
2.6 S
Transcribed Image Text:A place-kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 23.3 m/s at an angle of 53.5° to the horizontal. (a) By how much does the ball clear or fall short of clearing the crossbar? (Enter a negative answer if it falls short.) (b) Does the ball approach the crossbar while still rising or while falling? Step 1 (a) The football is moving with constant velocity in the x-direction. We denote the initial horizontal position of the football by xo and its initial x-component of velocity by Vox. For the position in the horizontal direction in terms of time t, we have x = xo + Voxt. Solving for t gives the following equation. t = We do not know the horizontal component of velocity but we can calculate it from the angle of the football's path to the horizontal. Remember that we use trigonometry to find horizontal and vertical components of a vector when we know its magnitude at a particular angle. We have 23.3 m/s cos 53.5 ✔✔✔ 1/s) co 53.5 13.86✔ Vox x-xo Vox t = = 23.3✔ Step 2 We are looking for the time for the ball to reach the goal, wherex - X is the distance to the goal. Substituting the value we found for Vox into the equation for t above, gives the following for t. 36 13.9 m) - (@ 0✔ 36 m 13.9 m/s 0 m = 2.59✔ 13.9 m/s. 2.6 S
Step 3
We will use the equation for the y-component of displacement under constant acceleration to find the height of the ball at this time, where yo is the initial vertical displacement, voy is the
initial y-component of velocity, and is the vertical acceleration.
y = Yo + voyt + 12/2a, t²
First we need to calculate voy Using trigonometry, we have
23.3✔
23.3 m/s sin 53.5✔
1/s) si
Step 4
Voy
=
ay
0 + 18.7
2.6
53.5
✔
Substituting the values we know into the equation for the y-component of displacement above, gives us the following result. Remember that ay = -g. Be careful with signs!
m/s)(
s) + ²/²/(1
y =
Your response differs from the correct answer by more than 10%. Double check your calculations. ²)
° = 18.7✔
18.7 m/s.
-9.81 ]✔ m/s²)( [
36
X
Your response differs from the correct answer by more than 100%. m
X
Transcribed Image Text:Step 3 We will use the equation for the y-component of displacement under constant acceleration to find the height of the ball at this time, where yo is the initial vertical displacement, voy is the initial y-component of velocity, and is the vertical acceleration. y = Yo + voyt + 12/2a, t² First we need to calculate voy Using trigonometry, we have 23.3✔ 23.3 m/s sin 53.5✔ 1/s) si Step 4 Voy = ay 0 + 18.7 2.6 53.5 ✔ Substituting the values we know into the equation for the y-component of displacement above, gives us the following result. Remember that ay = -g. Be careful with signs! m/s)( s) + ²/²/(1 y = Your response differs from the correct answer by more than 10%. Double check your calculations. ²) ° = 18.7✔ 18.7 m/s. -9.81 ]✔ m/s²)( [ 36 X Your response differs from the correct answer by more than 100%. m X
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