When a ball is thrown straight down from the top of a tall building, with initial velocity 30 ft/sec, the distance from the release point at time t in seconds is given by s(t)=16t^2+30t.
If the release point is 300 ft above ground, what is the velocity of the ball at the time it hits the ground?
I think you would have to average the feet per seconds but I'm not sure.
To find the velocity of the ball at the time it hits the ground, we need to first determine the time it takes for the ball to reach the ground. We know that the release point is 300 ft above the ground.
In the given equation, s(t) represents the distance from the release point at time t. When the ball hits the ground, the distance from the release point will be equal to zero. Therefore, we can set s(t) = 0 and solve for t.
0 = 16t^2 + 30t
To solve this quadratic equation, we can factor it or use the quadratic formula. However, in this case, since the equation doesn't factor easily, let's use the quadratic formula.
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 16, b = 30, and c = 0. Plugging in these values, we get:
t = (-30 ± √(30^2 - 4 * 16 * 0)) / (2 * 16)
t = (-30 ± 30) / 32
We take the positive value because time cannot be negative in this context:
t = (30 - 30) / 32 or t = (30 + 30) / 32
t = 0 / 32 or t = 60 / 32
t = 0 or t ≈ 1.875
Now that we have found the time it takes for the ball to hit the ground, we can find its velocity at that time.
To find the velocity, we can take the derivative of the position function s(t) with respect to time t. The derivative of s(t) will give us the instantaneous rate of change of the distance, which is the velocity.
s(t) = 16t^2 + 30t
To differentiate, we can use the power rule and the sum rule of derivatives:
s'(t) = 2 * 16t + 30
s'(t) = 32t + 30
Now, substitute t = 1.875 into the derivative function to find the velocity:
s'(1.875) = 32 * 1.875 + 30
s'(1.875) ≈ 60.625 ft/sec
Therefore, the velocity of the ball at the time it hits the ground is approximately 60.625 ft/sec.
Note: To solve the quadratic equation, we used the quadratic formula, but you mentioned averaging. Averaging would not be applicable in this case because we need to solve the equation precisely to determine the time and velocity.