Answer to Question #248409 in Calculus for Bman

Question

A bullet is fired in the air vertically from ground level with an initial velocity of 200 m/s. Find the bullet’s maximum velocity and maximum height.

Answer 1

In terms of this question windage is not taking into account.

According to the formula: $s(t)=s{\scriptscriptstyle 0}+v{\scriptscriptstyle 0}t-{\frac 1 2}gt^2$, $s{\scriptscriptstyle 0} = 0$, $v{\scriptscriptstyle 0} = 200$ m/s

$v(t) = v{\scriptscriptstyle 0} - gt$

We can see that with increasing of time the velocity is decreasing, so the maximum velocity will be at the moment when t = 0 and it is equal to 200 m/s(also, the exact same velocity can be seen at the moment when on the way back bullet hits the groung).

It is also known that bullet has the maximum height when v(t) = 0, so:

$v{\scriptscriptstyle 0} = gt$

$t =v{\scriptscriptstyle 0} *{\frac 1 g}= {\frac {200} {9.8}} = 20.4s$ - that is the time when the bullet will reach it's maximum height

By substitute t = 20.4 in the formula for height we get:

$s(20.4) = 0+200*20.4-{\frac 1 2}*9.8*416.16 = 2041 m$

The maximum velocity is 200m/s, the maximum height is 2041m