A rock is thrown off of a 100 foot cliff with an upward velocity of 45 m/s. As a result its height after t seconds is given by the formula: h(t)= 100+45t-5t^2
The height after 6 seconds is 190
What is the velocity after 6 seconds?
To find the velocity after 6 seconds, we need to find the derivative of the height function with respect to time. This derivative will give us the rate at which the height is changing with time, which is the velocity.
Given that the height function is h(t) = 100 + 45t - 5t^2, we can find the derivative by differentiating each term separately.
The derivative of the constant term (100) is 0 since a constant does not change with time.
The derivative of the linear term (45t) is 45 since the derivative of t is 1.
The derivative of the quadratic term (-5t^2) is -10t since the derivative of t^2 is 2t and we multiply by the coefficient.
Adding all these derivatives together, we get the derivative of the height function:
h'(t) = 0 + 45 - 10t = 45 - 10t
To find the velocity after 6 seconds, substitute t = 6 into the derivative:
h'(6) = 45 - 10(6) = 45 - 60 = -15
Therefore, the velocity after 6 seconds is -15 m/s.
To find the velocity after 6 seconds, we need to differentiate the equation for height with respect to time and evaluate it at t = 6 seconds.
The equation for height is given as h(t) = 100 + 45t - 5t^2.
To find the velocity, we differentiate h(t) with respect to t:
h'(t) = 45 - 10t
Now we substitute t = 6 into the expression for the velocity:
h'(6) = 45 - 10(6)
= 45 - 60
= -15
Therefore, the velocity after 6 seconds is -15 m/s.