velocity of a ball- if a ball is thrown into the air with a velocity of 40 ft/s, its height (in feet) after t seconds is given by y=40t-16t^2. find the velocity when t=2

v(t) = dy/dt

so find dy/dt, (that's a very easy one), then sub in t = 2

thank you

To find the velocity when t = 2, we need to differentiate the equation for height with respect to time (t) to get the velocity function.

The equation for height is given by:
y = 40t - 16t^2

To find the velocity function, we differentiate y with respect to t.
dy/dt = d(40t - 16t^2)/dt

The first term 40t differentiates to 40 (as the derivative of t with respect to t is 1).
The second term -16t^2 differentiates to -32t (using the power rule, where the derivative of t^n is n*t^(n-1)).

So, the velocity function is:
dy/dt = 40 - 32t

Now, substitute t = 2 into the velocity function to find the velocity when t = 2.
v = 40 - 32(2)
v = 40 - 64
v = -24 ft/s

Therefore, when t = 2, the velocity of the ball is -24 ft/s.