How do you solve this Calculus I problem
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How do you solve this Calculus I problem

[From: ] [author: ] [Date: 11-12-13] [Hit: ]
then its height in feet after t seconds is s(t) = 650 +32t - 16t^2.What will be the velocity of the ball when it hits the ground?What will be the acceleration of the ball when it hits the ground?-First of all, lets find the time that it takes for the ball to hit the ground, that is,......
I don't need the answer,,, I need the steps one would take to solve this problem. Broken down to their smallest possible steps.

If a ball is thrown into the air from the top of a building 650 feet above the ground with an initial velocity of 32 feet per second, then its height in feet after t seconds is s(t) = 650 +32t - 16t^2. What will be the velocity of the ball when it hits the ground? What will be the acceleration of the ball when it hits the ground?

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First of all, let's find the time that it takes for the ball to hit the ground, that is, when s = 0:
s(t) = 650 + 32t - 16t^2 = 0
t = (-32+-sqrt(32^2-4*650*-16))/(2*-16)
t = -5.45 or 7.45
Thus, t = 7.45s as t must be positive.

Now we need to find an equation for the velocity. The velocity is the rate of change of the position, relative to time; thus, it is the derivative of s(t):
v(t) = s'(t) = 32 - 2*16t = 32 - 32t
The velocity at t = 7.45 is:
v(7.45) = 32 - 32*7.45 = -206.46 ft/s or 206.46 ft/s downwards.

The acceleration is the rate of change of velocity:
a(t) = v'(t) = -32 ft/s^2

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When the ball touches ground velocity becomes zero since it comes to a halt
ie. ds/dt=v=0 differentiating and putting ds/dt=0 we get t=r sec and differentiating again we get
d^2s/dt^2=a=-32 ft/sec^2 since acc. Is negative therefore s is max when t=1 sec. Substituting this value in eq of s we get max S=650+16=666ft.
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