Question

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?

Answer 1

Answer: 469 feet

Step-by-step explanation:

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

`y=y_(o)+V_(o)t-(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)-(1)/(2)(32ft/s^(2))(6s)^(2)` (2)

Finally:

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