A stone is thrown straight up from the edge of a roof, 625 feet above the ground, at a speed of 20 feet per second.
A. Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 2 seconds later?
To determine the height of the stone 2 seconds later, we need to calculate its position using the equation of motion:
h(t) = h0 + v0t + (1/2)at^2
Where:
- h(t) is the height of the stone at time t
- h0 is the initial height (625 feet)
- v0 is the initial velocity (20 feet per second)
- t is the time elapsed (2 seconds)
- a is the acceleration due to gravity (-32 feet per second squared)
Plugging the given values into the equation, we can now calculate the height of the stone:
h(2) = 625 + (20 * 2) + (1/2)(-32)(2^2)
= 625 + 40 + (-16)
= 625 + 40 - 16
= 649 feet
Therefore, the stone will be at a height of 649 feet 2 seconds later.
h(t) = 625 + 20t - 16t^2
so, what is h(2)?