SOLUTION: Please help! Sorry it's more than one question
A ball is thrown straight up with an initial velocity of 40 m/s, so that its height (in meters) after t seconds is: h(t) = 2t(20 
Question 972110: Please help! Sorry it's more than one question
A ball is thrown straight up with an initial velocity of 40 m/s, so that its height (in meters) after t seconds is: h(t) = 2t(20 − t).
(a) Find an expression for the velocity of the ball at time t.
(b) What is the instantaneous velocity at t = 4? Is the ball rising or falling at this time?
(c) What is the maximum height achieved by the ball? Hint: the velocity of the ball will be zero when the ball is at its highest point.
(d) How fast will the ball be travelling when it hits the ground? Hint: first calculate time t when the ball hits the ground. Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! velocity(v) is defined as the first derivative of the position function
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h(t) = 2t(20 − t) = 40t - 2t^2
note that this is a parabola that opens downward
v = h'(t) = 40 - 4t
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a) v = h'(t) = 40 - 4t
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b) v = 40 - (4*4) = 24
ball is rising
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c) 0 = 40 - 4t
-4t = -40
t = 10 and
h(10) = 40t - 2t^2
h(10) = 400 - 200 = 200
max height is 200 m
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d) 0 = 40t - 2t^2
2t^2 = 40t
t = 20 and
h'(20) = 40 - 4*20
h'(20) = -40 which is a downward velocity of 40 m/s
