Calculus, Antiderivatives: A stone is thrown up at 18 ft/sec

C

chucknorrisfish

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A stone is thrown straight up from the edge of a roof 925 ft above the ground, at a speed of 18 ft per second. Acceleration is -32 ft/second squared.

A. How high is the stone 2 seconds later?
B. At what time does the stone hit the ground?
C. What is the velocity of the stone when it hits the ground.

so i know i need to find the antiDerivative, but i dont know what the equation is, or how to set it up. Please help me.
 
um...im not sure but ill try

so h= 925t - 32tsquared
h' = -64t + 925

a). --> -64(2) + 925
= 796?

b). the stone hits the ground when h(t)=0 so

h = - 32tsquared + 925t
0 = - 32tsquared + 925t? and then u solve and equate to 0 i think.

i hope i dont confuse u. [/url][/list]
 
Alright, i figures out A and B, maybe ya'll can help me with C.

a. the equation is -16(2)^2 + 925 + 18(2) gives you 897.
b. stone hits the ground at 8.1873, i just used quad formula on above equation.

c.?
 
From algebra, you have the basic equation for projectile motion:

. . . . .s(t) = -16t<sup>2</sup> + v<sub>0</sub>t + h<sub>0</sub>

...where s(t) is the height at time t seconds, v<sub>0</sub> is the initial velocity, and h<sub>0</sub> is the initial height. With calculus, of course, you can antidifferentiate twice to obtain the same equation.

. . . . .a(t) = -32

. . . . .v(t) = -32t + C
. . . . .v(0) = 0 + C = 18
. . . . .C = 18 = v<sub>0</sub>
. . . . .v(t) = -32t + 18

. . . . .s(t) = -16t<sup>2</sup> + 18t + D
. . . . .s(0) = 0 + 0 + D = 925
. . . . .D = 925 = h<sub>0</sub>

The rest of the exercise works largely as it did back in algebra:

A) Evaluate s(t) at t = 2.

B) Set s(t) equal to 0, and solve the quadratic for the time t.

C) Evaluate v(t) at the time t you found in (B).

Eliz.
 
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