The height, in feet, of a stone thrown vertically upward from 44 feet above ground level with an initial velocity of 40 ft/s is h = -16t^2 + 40t + 44. What is the velocity of the stone when it is 20 feet above the ground?
-16t^2 + 40t + 44 = 20
16 t^2 -40 t -24 = 0
2 t^2 -5 t - 3 = 0
( t-3)(2t+1) = 0
t = 3 seconds
v = dh/dt = -32 t+ 40 = -32(6)+40 = -192 + 40
To find the velocity of the stone when it is 20 feet above the ground, we need to find the derivative of the height function with respect to time.
Given: h = -16t^2 + 40t + 44
To find the derivative, we differentiate each term separately:
dh/dt = 2(-16)t^(2-1) + 40(t^(1-1)) + 0
dh/dt = -32t + 40
The derivative dh/dt represents the rate of change of height with respect to time, which is the velocity of the stone.
Now, substitute h = 20 into the derived equation and solve for t:
-32t + 40 = 20
-32t = -20
t = -20 / -32
t = 5/8
So, when the stone is 20 feet above the ground, the velocity of the stone is given by dh/dt = -32(5/8) + 40.
By simplifying the expression, we find:
dh/dt = -20 + 40
dh/dt = 20 ft/s
Therefore, the velocity of the stone when it is 20 feet above the ground is 20 ft/s.