Question

To start an avalanche on a mountain slope, an artillery W shell is fired with an initial velocity of$300\text{m}/\text{s}$at $55.0^o$above the horizontal. It explodes on the mountainside$42.0\text{s}$after firing. What are theandcoordinates of the shell where it explodes, relative to its firing point?

Short answer

$$\begin{gathered} x=7.23\times {10}^3\text{m} \\ y=1.68\times {10}^3\text{m} \end{gathered}$$

Step-by-step solution

 Given Data

Initial velocity = 300 m/s at 55.0^o

Definition of velocity.

The rate of change of an object's location with regard to a frame of reference is its velocity, which is a function of time. A statement of an object's speed and direction of motion is referred to as velocity.

Find the angle of projection.

To begin, use the following equation to get the shell's x-coordinate, which is its horizontal range:

$x=x_i+v_{xi}t+\frac{1}{2}{a}_x{t}^2$

Let's say the firing point is at $x_i=0$. In the horizontal direction, there is no acceleration $a_x=0$, and the starting velocity in the horizontal direction is $v_i\cos {\theta }_i$:

Then, using the following equation, get the shells y-coordinate: