Homework SourseA stone is thrown straight down from the edge of a roof 750
(A). Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 6 seconds later?
(B). At what time does the stone hit the ground?
(C). What is the velocity of the stone when it hits the ground?
Solution
Since acceleration is -32 feet per second squared, the equation for acceleration is:
A(t) = -32
To get the equation for velocity we intregrate the acceleration equation. The antiderivative is:
V(t) = -32t + C C is just the initial velocity, so that gives us
V(t) = -32t - 5
To get the equation for position, we integrate the velocity equation:
H(t) = -32t^2/2 - 5t + C where C is initial position
H(t) = -16t^2 - 5t + 750
Now that we have equations, we are ready to start answering the questions!!
(A) We just plug in 6 to our equation for position to get H(6) = -16*6^2 - 5*6 + 750 = 144 feet
(B) We use 0 as the value for value for position and then we solve for t:
0 = -16t^2 - 5t + 750
To solve this, we just use the quadratic formula. the approximate answer is 6.69206 seconds.
(C) To find this, we use the time that we just calculated and plug it into our equation for velocity.
V(6.69206) = -32(6.69206) - 5 = 209.14592
That\'s all!
