a diver jumps from a platform at time t=0 seconds. the distance of the diver above water level at time t is given by s(t)=-4.9t^2 +4.9t+10, where s is in meters
c.) find when velocity equal zero. find the maximum height of the diver
thank you
v(t) = -9.8t + 4.9
so, when is v=0?
Use that value of t to find s(t)
To find the time when the velocity of the diver is equal to zero, we need to find the derivative of the distance function and solve for when it is equal to zero. The derivative represents the rate of change of the function.
First, let's find the derivative of s(t) with respect to t:
s'(t) = -9.8t + 4.9
Next, let's set s'(t) equal to zero and solve for t:
-9.8t + 4.9 = 0
Adding 9.8t to both sides:
4.9 = 9.8t
Dividing both sides by 9.8:
t = 4.9 / 9.8
t = 0.5 seconds
Therefore, the velocity of the diver is equal to zero at t = 0.5 seconds.
To find the maximum height of the diver, we need to examine the vertex of the parabolic function s(t) = -4.9t^2 + 4.9t + 10. The vertex represents the highest or lowest point of the curve.
The formula for the x-coordinate of the vertex of a quadratic function in the form ax^2 + bx + c is given by x = -b / (2a).
In this case, the coefficient of t^2 is -4.9, the coefficient of t is 4.9, and the constant term is 10.
x = -4.9 / (2(-4.9)) = -0.5
To find the y-coordinate of the vertex, substitute the x-coordinate into the function:
s(-0.5) = -4.9(-0.5)^2 + 4.9(-0.5) + 10
s(-0.5) = -4.9(0.25) - 2.45 + 10
s(-0.5) = -1.225 - 2.45 + 10
s(-0.5) = 6.325
The maximum height of the diver is 6.325 meters.
Therefore, the time when the velocity equals zero is 0.5 seconds, and the maximum height of the diver is 6.325 meters.