Question

find the distance covered bt the bullet which is shot by a gun with an acceleration of 8 and with a velocity of 600 taking the gravity 10?

Answer 1

Task. Find the distance covered by the bullet which is shot by a gun with an acceleration of $a=8$ $m/s^{2}$ and with a velocity of $v_{0}=600$ $m/s$ taking the gravity $g=10$ $m/s^{2}$?

Solution. In fact there is not enough data to solve this problem. We also need the height $h$ at which the bullet was shot and the angle $\alpha$ between the initial velocity and the surface of earth.

Assume that the angle is zero so the bullet is shot parallel to the surface of earth.

There is a gravitation force acting on the bullet, and so its motion can be regarded as a sum of two motions: vertical and horizontal. Horizontal motion has initial velocity $v_{0}=600$ $m/s$ and constant acceleration $a=8$ $m/s$, so the distance covered by the bullet at time $t$ is given by the formula

$d(t)=v_{0}t+\frac{at^{2}}{2}.$

On the other hand, the vertical motion has zero initial velocity and constant acceleration $g=-10$ $m/s^{2}$. Let $h_{0}$ be the initial height of the bullet. Then its height at time $t$ is given by the formula:

$h(t)=h_{0}-\frac{gt^{2}}{2}.$

Therefore the bullet will fall to the ground at time $t$ such that

$h(t)=h_{0}-\frac{gt^{2}}{2}=0,$

whence

$\bar{t}=\sqrt{\frac{2h_{0}}{g}}.$

Then the distance covered by the bullet can be obtained by substituting $\bar{t}$ into the formula for $d$:

$d(\bar{t})=v_{0}\bar{t}+\frac{a\bar{t}^{2}}{2},$

so

$d(\bar{t})=v_{0}\bar{t}+\frac{a\bar{t}^{2}}{2}=v_{0}\sqrt{\frac{2h_{0}}{g}}+\frac{a}{2}\cdot\frac{2h_{0}}{g}=v_{0}\sqrt{\frac{2h_{0}}{g}}+\frac{h_{0}a}{g}.$

Answer. There is not enough data for solving the problem. However is we assume that the bullet is shot parallel to earth surface at height $h_{0}$, then it will cover the distance

$v_{0}\sqrt{\frac{2h_{0}}{g}}+\frac{h_{0}a}{g}.$