What is the initial velocity of an object thrown directly upward at a height of s= -16t^2 + 48t + 256 feet ?

Question

after t seconds? and how do you find the answer?

Answer 1

you simply take the derivative

s= -16t^2 + 48t + 256 feet is the position equation

s'= -32t + 48 is the velocity equation

s'' = -32 is the acceleration

so your answer would be s'(t)= -32t + 48

Answer 2

Let's pretend you plotted position versus time on a graph. Using Calculus concepts, you know that in this type of graph, slope of the line at a specific point on the function is given as velocity that the object is traveling (m/s).

Therefore, for any specific point in time, the derivative of the function gives the function equation for the velocity at that position and point in time.

So, velocity is given as s(t) = -32t + 48. Assuming that the initial velocity happens at t = 0, it is s(0) = -32(0) + 48. The initial velocity is 48 units / second.

After t seconds, it is s(t) = -32t + 48 units / second. Therefore, you plug in any specific time you want into the function to find the particular velocity at that time.

As a side note, when the answer is positive, that means that the object is traveling upwards, assuming you choose upwards as positive. If you should get a negative answer, that now means the object is now falling with the given solved speed.