The height in feet above the ground of a ball thrown upwards from the top of a building is given by s=-16t^2 + 160t + 200, where t is the time in seconds.
If the maximum height is 600 feet, what is v^-1(32)?
The answer is supposed to be 4 seconds, but I don't understand how. please explain
Since velocity is the derivative of position,
v(t) = -32t + 160
now just solve
-32t+160 = 32
-32t+128 = 0
t = 4
Thank you very much!
To solve this problem, we need to find the value of t when the height of the ball is 32 feet.
Given:
s(t) = -16t^2 + 160t + 200
s(t) represents the height of the ball in feet at time t.
We know that the maximum height is 600 feet. This occurs at the vertex of the parabolic equation. The vertex can be found using this formula:
t_vertex = -b/2a
In our case, a = -16 and b = 160. Substituting these values into the formula, we get:
t_vertex = -(160)/(2*(-16))
= -160/-32
= 5
So, the time at which the ball reaches its maximum height is t_vertex = 5 seconds.
Now we need to find the value of t when the height of the ball is 32 feet. We can set up an equation:
-16t^2 + 160t + 200 = 32
Simplifying this equation, we get:
-16t^2 + 160t + 200 - 32 = 0
-16t^2 + 160t + 168 = 0
To solve this quadratic equation, we can apply the quadratic formula:
t = (-b±√(b^2-4ac))/(2a)
Here, a = -16, b = 160, and c = 168. Substituting these values into the quadratic formula, we get:
t = (-160±√(160^2-4*(-16)*168))/(2*(-16))
= (-160±√(25600+10752))/(-32)
= (-160±√(36352))/(-32)
Since we are looking for the positive value of t, we can disregard the negative result from the ± operation. Simplifying further, we have:
t = (-160+√(36352))/(-32)
Evaluating this expression using a calculator, we find that t ≈ -0.42.
Since time cannot be negative in this context, we disregard this result.
Therefore, the value of t when the height of the ball is 32 feet is approximately 4 seconds, as stated in the problem.
To find v^-1(32), we need to determine the time at which the velocity of the ball is 32 feet per second.
The velocity of the ball at time t is given by the derivative of the height function s(t) with respect to time. So, let's first find the derivative of s(t):
s(t) = -16t^2 + 160t + 200
Taking the derivative of s(t) with respect to t, we get:
s'(t) = -32t + 160
Now we can set s'(t) equal to 32 and solve for t:
-32t + 160 = 32
Starting with -32t + 160 = 32, we can simplify the equation:
-32t = -128
Dividing both sides by -32, we find:
t = 4
Therefore, v^-1(32) = 4 seconds. The ball reaches a velocity of 32 feet per second after 4 seconds.