Ball 32
Ball 32
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a Ball is thrown vertically upward from the ground with an initial velocity of 60 feet per second. How high w
a Ball is thrown vertically upward from the ground with an initial velocity of 60 feet per second. How high will the ball go?
a(t)=32 feet per second acceleration due to gravity
Remember that acceleration is the derivative of velocity, and velocity is the derivative of distance. So there are two ways these problems are usually done-- either you derive the equations yourself by taking derivatives/integrating, or your teacher gives you the formulas in advance.
If you want to derive them, given that a(t) = -32 (negative because it goes down):
v(t) = -32t + v_0 -- just integrate a(t) with respect to t, and v_0 is the initial velocity
s(t) = -16t^2 + v_0t + s_0 -- integrating v(t) with respect to t, s_0 is initial height.
Now fill everything in given the information from the problem:
thrown "from the ground" = initial height is 0 (ground is always considered height 0)
initial velocity = 60 ft/sec = v_0 (positive because it is thrown upwards)
plug into s(t) to get:
s(t) = -16t^2 + 60t
now you want to find the maximum height... so to do this, you use usual min/max methods from calculus, or you can use the fact that it's a parabola and find the vertex using methods from precalc. I don't know which class this is for, so I'll sketch both methods:
calc method: find the max for s(t) doing the usual method of taking the derivative and setting it equal to 0
s'(t) = -32t + 60
0 = -32t + 60
60 = 32t
60/32 = t
verify that this is a max
precalc method: complete the square to find the vertex of the parabola (it will be downward pointing) and this corresponds to the max
actually, notice that in the calc method, when you take the derivative, you're just getting the velocity back again. this makes sense because when the ball reaches its maximum height is the same moment that the velocity is zero (the velocity is initially 60, upward, but gravity is acting against it to slow the ball's upward movement until finally the velocity is 0 for a split-second, then the velocity is negative as it travels downward towards the ground).
so now that I think about it, you could probably just say this instead of going through the process of finding the maximum... depends on which class this is for.