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?

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cryssatemp cryssatemp · Answer 1 · 2019-07-04T01:45:51+02:00

Explanation:

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

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

Where:

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

[tex]y_{o}=925ft[/tex] is the initial height of the stone

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

[tex]t=6s[/tex] is the time at which we need to find the height

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

Having this clear, let's find [tex]y[/tex] from (1):

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

Finally:

[tex]y=469ft[/tex] This is the height of the stone at t=6s