h = v{sin 8;)t –g(ty,max), which using ty,max gave her n = v (sin )(ain .) - san .)*. simpliying this, she got h =, v?(sin? 0) _ v(sin? 0,). which 2g reduced to this result. Part 4 of 9 - Maximum Height Abel is not sure why, from h = v? sin2 0, the height the tennis ball reaches is maximum when 0, = 90°, and 2g asks Kato to explain. Which of Kato's responses is correct? O "When 0, = 90°, sin? 0, = sin(20,) = 1, which is its maximum value, so h is maximum." "When 0, = 90°, sin? e, is maximum, so h is maximum." "When e, = 90°, sin 0, is minimum, soh is maximum."

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Please only answer part 4 of 9.

acceleration; therefore, the height is greatest here."
"For this 90° angle, the horizontal component of velocity is zero, so the initial velocity is
completely in the vertical direction, resulting in greatest height."
"For this angle, the vertical component of velocity is zero, so the initial velocity is purely in the
horizontal direction."
Abel is correct. As Abel recognized, the initial velocity is entirely in the vertical direction for a 90° launch, so the
height of the tennis ball is greatest for a launch angle of 90°.
Part 3 of 9 - Maximum Height
Abel challenges Kato to use the equation for the time for any projectile to reach maximum height in the
absence of air resistance, given by tymax = V, sin o, and the kinematic equation y, = Y, + v,(sin 0pt -gt,
to determine the maximum height, h, of the tennis ball, given the initial launch angle and speed. Which of
Kato's attempts is correct?
v? sin? 0,
| Kato tries substituting tymax for t, 0 for Y, and h for yp and gets h =
2g
Kato says that it is not at all possible to determine the maximum height of a tennis ball using these
equations.
Kato tries substituting 2ty,max
v? sin? o,
for t, 0 for y, and h for y, and gets h =
v,? sin? o,
Kato tries substituting ty,max for t, 0 for y, and h for y, and gets h =
Kato is correct. This equation can be used to determine the maximum height of a tennis ball in the absence of air
resistance, given the initial velocity and launch angle. When Kato made her substitutions, she got
h = v,(sin ê)t - gty.max)², which using ty,max = V; sind, gave her
h= v(sin e)( sin )-, sin 0,)°. simplifying this, she got h =
("; sin
v?(sin? 0) _ v?(sin? 0),
which
2g
reduced to this result.
Part 4 of 9 - Maximum Height
Abel is not sure why, from h = Vi sin- ",, the height the tennis ball reaches is maximum when 0, = 90°, and
29
asks Kato to explain. Which of Kato's responses is correct?
O "When 0, = 90°, sin? 0, = sin(20,) = 1, which is its maximum value, so h is maximum."
O "When 0, = 90°, sin? 0, is maximum, so h is maximum."
"When 0, = 90°, sin? 0, is minimum, so h is maximum."
"When 0, = 90°, sin 0, = 1, sin? 0, = 2·1 = 2, which is its maximum value, so that means h is
%3D
maximum."
Transcribed Image Text:acceleration; therefore, the height is greatest here." "For this 90° angle, the horizontal component of velocity is zero, so the initial velocity is completely in the vertical direction, resulting in greatest height." "For this angle, the vertical component of velocity is zero, so the initial velocity is purely in the horizontal direction." Abel is correct. As Abel recognized, the initial velocity is entirely in the vertical direction for a 90° launch, so the height of the tennis ball is greatest for a launch angle of 90°. Part 3 of 9 - Maximum Height Abel challenges Kato to use the equation for the time for any projectile to reach maximum height in the absence of air resistance, given by tymax = V, sin o, and the kinematic equation y, = Y, + v,(sin 0pt -gt, to determine the maximum height, h, of the tennis ball, given the initial launch angle and speed. Which of Kato's attempts is correct? v? sin? 0, | Kato tries substituting tymax for t, 0 for Y, and h for yp and gets h = 2g Kato says that it is not at all possible to determine the maximum height of a tennis ball using these equations. Kato tries substituting 2ty,max v? sin? o, for t, 0 for y, and h for y, and gets h = v,? sin? o, Kato tries substituting ty,max for t, 0 for y, and h for y, and gets h = Kato is correct. This equation can be used to determine the maximum height of a tennis ball in the absence of air resistance, given the initial velocity and launch angle. When Kato made her substitutions, she got h = v,(sin ê)t - gty.max)², which using ty,max = V; sind, gave her h= v(sin e)( sin )-, sin 0,)°. simplifying this, she got h = ("; sin v?(sin? 0) _ v?(sin? 0), which 2g reduced to this result. Part 4 of 9 - Maximum Height Abel is not sure why, from h = Vi sin- ",, the height the tennis ball reaches is maximum when 0, = 90°, and 29 asks Kato to explain. Which of Kato's responses is correct? O "When 0, = 90°, sin? 0, = sin(20,) = 1, which is its maximum value, so h is maximum." O "When 0, = 90°, sin? 0, is maximum, so h is maximum." "When 0, = 90°, sin? 0, is minimum, so h is maximum." "When 0, = 90°, sin 0, = 1, sin? 0, = 2·1 = 2, which is its maximum value, so that means h is %3D maximum."
Kato and Abel are tennis partners, and are also in the same physics class. They are playing around with the
tennis ball machine, which launches tennis balls at an initial speed of 30 m/s. Kato wonders how high and how
far the tennis ball can go, and the pair decide to perform several experiments to find out.
This simulation shows a projectile launched from the origin. The initial speed of the projectile is 30 m/s. Kato
and Abel can use the slider to set the launch angle or enter an angle value by clicking inside the degrees box
and then clicking "fire" to set the projectile in motion. They can move the mouse pointer to see readouts of
the x- and y-coordinates for any point on the curve. They can click "pause" to stop the simulation at any point
and "clear" to remove traces of the projectile paths. Note that in this simulation, air resistance is ignored.
y (m)
75
zy =30 m/s
50
25
* (m)
25
50
75
100
125
Launch Angle
50
degrees
90
75
- 60
45
30
15
Time =
0.00 s
fire
pause
clear
Click here to open the simulation in a new window.
Part 1 of 9 - Maximum Height
Kato runs the pitching machine for several different angles, and she and Abel watch carefully to see what
angle results in the greatest height. What angle do they find?
| 90°
45°
15°
30°
They are correct. When the launch angle is 90°, the projectile reaches its greatest height.
Transcribed Image Text:Kato and Abel are tennis partners, and are also in the same physics class. They are playing around with the tennis ball machine, which launches tennis balls at an initial speed of 30 m/s. Kato wonders how high and how far the tennis ball can go, and the pair decide to perform several experiments to find out. This simulation shows a projectile launched from the origin. The initial speed of the projectile is 30 m/s. Kato and Abel can use the slider to set the launch angle or enter an angle value by clicking inside the degrees box and then clicking "fire" to set the projectile in motion. They can move the mouse pointer to see readouts of the x- and y-coordinates for any point on the curve. They can click "pause" to stop the simulation at any point and "clear" to remove traces of the projectile paths. Note that in this simulation, air resistance is ignored. y (m) 75 zy =30 m/s 50 25 * (m) 25 50 75 100 125 Launch Angle 50 degrees 90 75 - 60 45 30 15 Time = 0.00 s fire pause clear Click here to open the simulation in a new window. Part 1 of 9 - Maximum Height Kato runs the pitching machine for several different angles, and she and Abel watch carefully to see what angle results in the greatest height. What angle do they find? | 90° 45° 15° 30° They are correct. When the launch angle is 90°, the projectile reaches its greatest height.
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