calculus word problem with velocity?

Question

On the surface of Mars, a stone is tossed upward with an initial velocity of 40 ft/sec from a height of 6 feet above the surface of the planet. Its height is given by the function H(t)= -6.5t^2 +40t+6, where t is in seconds. How many seconds after the stone is released will it reach its maximum height? What is the stone's maximum height?

Answer 1

this is a simple dervitive problem....we recall that given a position the first derivitive would give a velocity at any time t, and a second derivitive would give an accelaration at any time t

so we need to take the derivitive of h(t)

so h(t) = -6.5t^2 +40t+6
h'(t)= -13 t +40

now you find when h'(t) is zero...this will tell you when the slope of a tangent line of position is zero...in otherwords...a max or min

so you find h'(t)=0=-13t+40
so t= 40/13

now we need the second derivitive to apply the second derivitive test (recall this tells us that if the h''(t) is negative at a specific t value we have a max, and if its positive at that t value its a min)

so h''(t)=-13, so for all values of t h''(t) is negative so at t=40 h'(t) is zero and h''(t) is negative so there is a max height at t= 40/13

now to find the height you plug 40/13 into h(t)

so h(40/13) i will leave to you cause this is teh easy part

hope this helps

matttlocke

Answer 2

Velocity = dx/dt
Acceleration = dv/dt

Velocity = acceleration * time

Time = velocity / acceleration = 40 / 13 = 3.076923 seconds (keeping more intermediate digits than are actually significant)

Distance = 1/2 * acceleration * time^2 = 6.5 * t^2 = 62 feet above launch elevation

Height = 62 + 6 = 68 feet

Answer 3

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Answer 4

The initial velocity, 40, and height, 6 are correctly built into the equation, so now all we need is the point in time of the maximum height.

Use local maxima rule, take the first derivative of the equation and solve for zero:

H'(t) = -13t + 40 = 0, solving for t you get t=3.08 seconds (rounded to nearest hundredth).

Going back to the original equation for height, plug 3.08 into the equation:

H(3.08) = -6.5(3.08)^2 +40(3.08) + 6 .... I'll let you take it from here.

Answer 5

find H'(t) and set it = to zero
thatll give you at what time (t) h(t) is maximized
then take that value and plug it back into H(t) and thats the max height