Asked by Dee Dee
A pair of glasses are dropped from the top of a 32.0 m high stadium. A pen is dropped from the same position 2.00 s later. How long does it take for the glasses to hit the ground? How high above the ground is the pen when the glasses hit the ground? (Disregard air resistance)
Answers
Answered by
GK
Find the time of fall by using:
y = (1/2)gt^2
(y is given and g = 9.8m/s^2, Solve for t)
When the glasses hit the ground, the pen has been falling for (t-2.00s). The height above the ground of the glasses is:
32.0m - (1/2)g(t-2)^2
y = (1/2)gt^2
(y is given and g = 9.8m/s^2, Solve for t)
When the glasses hit the ground, the pen has been falling for (t-2.00s). The height above the ground of the glasses is:
32.0m - (1/2)g(t-2)^2
Answered by
GK
Correction to the previous answer:
"The height above the ground of the <b>glasses</b> is:
32.0m - (1/2)g(t-2)^2" should read:
The height above the ground of the <b>pen</b> is:
32.0m - (1/2)g(t-2)^2
"The height above the ground of the <b>glasses</b> is:
32.0m - (1/2)g(t-2)^2" should read:
The height above the ground of the <b>pen</b> is:
32.0m - (1/2)g(t-2)^2
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