Kato asks why a 45° launch angle results in the maximum horizontal range. Abel uses R = 2V; sin 0, cos 0, to attempt an explanation. Which explanation is correct? |"I used the trigonometric identity sin(20) = 2 sin 0 cos 0 to rewrite the expression for the range, R = 2v? sin 0, cos 0, and got R = vf sin(20,). And, of course, when 0, = 45°, 20, = 90°, and sin 90° = 1, which is its maximum value." "Because sin 45° = cos 45° = 1, the maximum value for sine and cosine, R, is maximum when 0, = 45°." "I used the trigonometric identity, sin(20) = sin 0 cos 0, to rewrite the expression for the range, 2v? sin 0, cos 0, and got R = "² sin(20). When 0, = 45°, 20, = 90°, and sin 90° = 1, R = which is its maximum value." "This cannot be shown using an equation. The results are empirical, and we only know this from direct observation."
Kato asks why a 45° launch angle results in the maximum horizontal range. Abel uses R = 2V; sin 0, cos 0, to attempt an explanation. Which explanation is correct? |"I used the trigonometric identity sin(20) = 2 sin 0 cos 0 to rewrite the expression for the range, R = 2v? sin 0, cos 0, and got R = vf sin(20,). And, of course, when 0, = 45°, 20, = 90°, and sin 90° = 1, which is its maximum value." "Because sin 45° = cos 45° = 1, the maximum value for sine and cosine, R, is maximum when 0, = 45°." "I used the trigonometric identity, sin(20) = sin 0 cos 0, to rewrite the expression for the range, 2v? sin 0, cos 0, and got R = "² sin(20). When 0, = 45°, 20, = 90°, and sin 90° = 1, R = which is its maximum value." "This cannot be shown using an equation. The results are empirical, and we only know this from direct observation."
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Transcribed Image Text:2v2 sin 0, cos 0;
Kato asks why a 45° launch angle results in the maximum horizontal range. Abel uses R =
to attempt an explanation. Which explanation is correct?
|"I used the trigonometric identity sin(20) = 2 sin 0 cos 0 to rewrite the expression for the
2v, sin 0, cos 0,, and got R = V sin(20,). And, of course, when 0, = 45°, 20, = 90°,
range, R =
and sin 90° = 1, which is its maximum value."
"Because sin 45° = cos 45° = 1, the maximum value for sine and cosine, R, is maximum when
0, = 45°."
"I used the trigonometric identity, sin(20) = sin 0 cos 0, to rewrite the expression for the range,
2v? sin 0, cos , and got R =
ŕ sin(20). When 0, = 45°, 20, = 90°, and sin 90° = 1,
R =
which is its maximum value."
"This cannot be shown using an equation. The results are empirical, and we only know this from
direct observation."
Abel is correct. He knows that for angles greater than 45°, although the total time of flight of the projectile is longer,
the horizontal component of velocity is less, so its horizontal displacement is less. For angles less than 45°, although
the horizontal component of velocity is greater, it requires less time to make the vertical component of the velocity
zero, so the total time of flight of the projectile is shorter, and its horizontal displacement is also less than at a 45°
launch angle.
Part 9 of 9 - Analysis
Now Abel and Kato use what they learned to answer the following problem.
The initial speed of a tennis ball is 57.5 m/s and the launch angle is 0, = 16°. Neglect air resistance.
What is the maximum height, h, of the tennis ball?
m
What is the range, R, of the tennis ball?
m
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