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omeli [17]
3 years ago
6

A stone is thrown straight up from the edge of a roof, 925 feet above the ground, at a speed of 20 feet per second. Remembering

that the acceleration due to gravity is -32 feet per second squared, how high is the stone 6 seconds later?
Physics
1 answer:
Black_prince [1.1K]3 years ago
4 0
<h2>Answer: 469 feet</h2>

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

y=y_{o}+V_{o}t-\frac{1}{2}gt^{2} (1)

Where:

y is the height of the stone at 6s (the value we want to find)

y_{o}=925ft is the initial height of the stone

V_{o}=20ft/s is the initial velocity of the stone

t=6s is the time  at which we need to find the height

g=32ft/s^{2} is the acceleration due to gravity

Having this clear, let's find y from (1):

y=925ft+(20ft/s)(6s)-\frac{1}{2}(32ft/s^{2})(6s)^{2} (2)

Finally:

y=469ft This is the height of the stone at t=6s

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