A 45- caliber bullet is fired straight up from the surface of the moon with an i

Question

A 45- caliber bullet is fired straight up from the surface of the moon with an initial velocity of 961 feet per second . The function that describes the height (in feet) of the bullet at any time , t, is : s(t) = 961(t)-2.6(t)^2 . On Earth , in the absence of air resistance , the height (in feet) of the bullet at any time , t, is given by the function : s(t) = 961(t) -16(t)^2 .

a. how long will the bullet be in the air in both cases, before it come back and hits the ground?

b. when will the bullet reach it's highest point in each case and what is that maximum height ?

c. when  the bullet hits the ground , what would the velocity be in both cases?

Explanation / Answer

a)set both equal to 0 (this means position is 0) so you should expect one of the answers to be 0

t(961-2.6t)=0

t=0

t=369.6 seconds

and the same for the second returns

t=0

t=60.06 seconds

b) the bullet will reach it's highest point when the time is 1/2 as that is when it returns back to the ground. You know the total times from the section before so

961(369.6/2)-2.6(369.6^2/2)= 88800.096 feet

and 961(60.06/2)-16(60.06^2/2) =14430.01 feet

c)the given equations are for postion so the derivative will be velocity equations

s'=961t-2.6t^2 d/dt = 961-5.2t (using the time of flight from "a" we get -960.92 ft/s meaning the bullet is now traveling in the opposite direction from the begining so 960.92 ft/s down the same done for the second equation results in the same velocity.

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