##### Solve height speed function in calculus

 Calculus Tutor: None Selected Time limit: 1 Day

Apr 16th, 2015

b. Deriving the position function (height), you get the rate of change of the height, the vertical velocity function. At all times past t=0, the upward velocity will be decreasing because of the acceleration due to gravity. At t=5, the object is still traveling upwards at 40 ft/sec.

c. Finding the critical number, you get the time at which the height stops changing for a moment in time, dt. Since the object is moving up then moving down, it will be at its maximum height at the time at which it changes from positive to negative velocity. At this moment in time, the object is motionless. Thereafter, it begins traveling downward.

d. Graphically, with constant acceleration, position changes according to a symmetric parabola (if acceleration opposes velocity at t=0). This means that the time it takes for the object to reach max height is equal to the time it takes for it to fall back down to initial height. So from its launch time, it takes 2(critical number) to fall back down.

e. Graphically, the magnitude of velocity, given constant acceleration, is the same when it starts moving up at initial height to when it hits the ground at the same height. The direction of travel is opposite, so the sign is opposite. v(i) = - v(f) of a projectile launched straight up.

Note: Interpret most of this only for this example. Some of the statements are specific to upwardly launched projectiles.

Apr 16th, 2015

So from

a.  h(t) = -16t^2 + 200t  how did b. become -32t + 200.

Apr 16th, 2015

omg I was confused at how you arrived at -32t + 200 forgetting to apply

d/dt ((-16)t^2)

= 2(-16)t^2-1

= -32t^1

= -32t

Apr 16th, 2015

Yep, that's the derivation with respect to time of the position function with your chosen velocity. -32t means the velocity decreases 32 units per time unit.

Apr 17th, 2015

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Apr 16th, 2015
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Apr 16th, 2015
Dec 9th, 2016
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