## University Calculus: Early Transcendentals (3rd Edition)

a) On the Moon, the bullet is aloft for $320$ seconds and goes as high as $66560$ feet. b) On Earth, the bullet is aloft for $52$ seconds and goes as high as $10816$ feet.
a) On the Moon: $$s(t)=832t-2.6t^2(ft)$$ - To find how long the bullet is aloft, we look for 2 values of $t$, one corresponding to the time the bullet was shot and the other to the time the bullet went back to the ground, both having $s(t)=0$. $$s(t)=0$$ $$832t-2.6t^2=0$$ $$t(832-2.6t)=0$$ $$t=0sec\hspace{1cm}\text{or}\hspace{1cm}t=320sec$$ - $t=0$ corresponds to the ground-leaving moment of the bullet. - $t=320$ corresponds to the ground-getting back moment of the bullet. So the bullet is aloft for $320$ seconds. - Find $v(t)$: $$v(t)=\frac{ds}{dt}=(832t-2.6t^2)'=832-5.2t$$ So $v(t)=0$ when $832-5.2t=0$, or $t=160sec$. The moment the velocity of the bullet is $0$ is the moment it reaches its maximum height, which is $160$ seconds after it was shot. The height the bullet went is $$s(160)=832\times160-2.6\times160^2=66560(ft)$$ b) On Earth: $$s(t)=832t-16t^2(ft)$$ - Again, to find how long the bullet is aloft, we look for 2 values of $t$, one corresponding to the time the bullet was shot and the other to the time the bullet went back to the ground, both having $s(t)=0$. $$s(t)=0$$ $$832t-16t^2=0$$ $$t(832-16t)=0$$ $$t=0sec\hspace{1cm}\text{or}\hspace{1cm}t=52sec$$ - $t=0$ corresponds to the ground-leaving moment of the bullet. - $t=52$ corresponds to the ground-getting back moment of the bullet. So the bullet is aloft for $52$ seconds. - Find $v(t)$: $$v(t)=\frac{ds}{dt}=(832t-16t^2)'=832-32t$$ So $v(t)=0$ when $832-32t=0$, or $t=26sec$. The moment the velocity of the bullet is $0$ is the moment it reaches its maximum height, which is $26$ seconds after it was shot. So the height the bullet went is $$s(26)=832\times26-16\times26^2=10816(ft)$$