mykamorrow
07.10.2020 •
Physics
A heavy rock is shot upward from the edge of a vertical cliff. It leaves the edge of the cliff with an initial velocity of 12 m/s directed at 26° from the vertical and experiences no appreciable air resistance as it travels. The distance the rock landed was 40.4 m away from the cliff. How high is the cliff?
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Ответ:
The height of the building is 88.63 m.
Explanation:
Given;
initial component of vertical velocity, = 12 m/s sin 26° = 5.26 m/s
initial horizontal component of the velocity, = 12 m/s cos 26° =10.786 m/s
horizontal distance traveled by the rock, x = 40.4 m
time of flight is calculated as;
x = t
t = x /
t = 40.4 / 10.786
t = 3.75 s
Determine the final vertical velocity of the ball;
Determine the height of the rock;
Therefore, the height of the building is 88.63 m.
Ответ: