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?
h(6) = 100+45(6) - 5(36) = 190 m
Wow, this person was able to throw a rock upwards 90 m, the height of roughly a football field ??
velocity = h' (t) = 45 - 10t
when t=6
velocity = 45 - 60 = -15 m/sec
Well, after 6 seconds, the velocity of the rock would be the derivative of the height function at that time. Let me crunch some numbers for you... *beep boop beep*
Using the power rule for differentiation, we find that the derivative of h(t) = 100 + 45t - 5t^2 is h'(t) = 45 - 10t.
Now, plugging in t = 6 seconds, we get h'(6) = 45 - 10(6) = 45 - 60 = -15 m/s.
So, the velocity after 6 seconds is -15 m/s. The rock seems to be taking a slight tumble there!
To find the velocity after 6 seconds, we need to take the derivative of the height function with respect to time.
First, let's differentiate the function h(t)=100+45t-5t^2:
h'(t) = 0 + 45 - 10t
Now let's plug in t = 6:
h'(6) = 45 - 10(6) = 45 - 60 = -15
The velocity after 6 seconds is -15 m/s (negative sign indicates that the velocity is downward).
To find the velocity after 6 seconds, we need to find the derivative of the height function with respect to time.
The height function is given by: h(t) = 100 + 45t - 5t^2
Taking the derivative of this function will give us the rate of change of height with respect to time, which is the velocity.
Taking the derivative of the height function, we get:
h'(t) = 45 - 10t
Now we can plug in the value t = 6 into the derivative function to find the velocity after 6 seconds:
h'(6) = 45 - 10(6)
= 45 - 60
= -15
Therefore, the velocity after 6 seconds is -15 m/s.