PHYSICS HELP ASAP PROJECTILE MOTION PLEASE!?!?!?!?
2020-08-26 10:21 pm
During eruptions, chunks of solid rock can be blasted out of the volcano; these projectiles are called volcanic bombs. Suppose a bomb is ejected at 85.0 m/s, 28.0° above the horizontal as shown. If a bomb lands on the ground 22.03 seconds after it is launched how high is the mouth of the volcano? How far, horizontally, from the mouth of the volcano does the bomb land?
回答 (1)
2020-08-26 10:58 pm
Take g = 9.81 m/s²
Take all upward quantities as positive, and downward quantities as negative.
Refer to the figure below.
Consider the vertical motion (uniform acceleration motion):
Initial velocity, u(y) = 85.0 sin28° m/s
Acceleration, a(y) = -9.81 m/s²
Time taken, t = 22.03 s
s(y) = u(y) t + (1/2) a(y) t²
s(y) = (85.0 sin28°) (22.03) + (1/2)(-9.81)(22.03)²
s(y) = -1500 m
Consider the horizontal motion (uniform velocity motion):
s(x) = v(x) t = (85.0 cos28°) (22.03) = 1650 m
The answer:
The height of the mouth of the volcano = 1500 m
The distance that the bomb land from the mouth of the volcano = 1650 m
Take all upward quantities as positive, and downward quantities as negative.
Refer to the figure below.
Consider the vertical motion (uniform acceleration motion):
Initial velocity, u(y) = 85.0 sin28° m/s
Acceleration, a(y) = -9.81 m/s²
Time taken, t = 22.03 s
s(y) = u(y) t + (1/2) a(y) t²
s(y) = (85.0 sin28°) (22.03) + (1/2)(-9.81)(22.03)²
s(y) = -1500 m
Consider the horizontal motion (uniform velocity motion):
s(x) = v(x) t = (85.0 cos28°) (22.03) = 1650 m
The answer:
The height of the mouth of the volcano = 1500 m
The distance that the bomb land from the mouth of the volcano = 1650 m
收錄日期: 2021-05-01 22:37:43
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