pleasw help A ball is thrown upward from a height of 432 feet above the ground,

Question

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A ball is thrown upward from a height of 432 feet above the ground, with an initial velocity of 96 feet per second. From physics it is known that the velocity at time t is v(t) = 96 - 32t feet per second. Find s(t): the function giving the height of the ball at time t. How long will the ball take to reach the ground? How high will the ball go? For the function f, given below, find the anti derivative F that satisfies F(1) = 0. f(x) = x6 - 2x - 3 - 3 The anti derivative that satisfies the given condition is F(x) =

Explanation / Answer

1.) s(t) = integral of v(t) since s'(t) = v(t)

s(t) = 96t-16t^2+c where c is integration constant

s(0) = 432, the ball is at 432 feet at t=0

=>c=432

=> s(t) = 96t-16t^2+432

For the ball to reach ground, s(t)=0

=> 96t-16t^2+432=0

=> t = 9 seconds

Max height of ball is when velocity of ball becomes zero and starts to fall down

=> v(t) = 0

96-32t=0=> t=3 seconds

s(t)=(96*3)-(16*3*3)+432=576 meters

2.)Anti derivative of f(x) = x^7/7 + x^-2 -3x+c

This function becomes 0 at x=1

=> 1/7+1-3+c=0 => c=13/7

Anti derivative of f(x) = x^7/7 + x^-2 -3x+13/7

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