Question

Please explain the process and equation to get the answer

Answer 1

The shell's horizontal position $x$ and vertical position $y$ are given by

$x=v_0\cos55.0^\circ\,t$

$y=v_0\sin55.0^\circ\,t-\dfrac g2t^2$

where $v_0$ is the given speed of 1.70 x 10^4 m/s, and $g$ is the acceleration due to gravity (taken here to be 9.80 m/s^2).

To find the horizontal range, you can use the range formula,

$x_{\rm max}=\dfrac{{v_0}^2\sin2\theta}g$

with $\theta$ being the angle at which the shell is fired.

Alternatively, we can work backwards and deal with part (b) first:

(b) The time spent in the air is the time it takes for the shell to reach the ground. To find that, you solve for $t$ in $y=0$:

$v_0\sin55.0^\circ\,t-\dfrac g2t^2=0\implies t\approx284\,\rm s$

(a) After this time, the shell will have traveled horizontally

$x=v_0\cos55.0^\circ(284\,\rm s)\approx2.77\times10^5\,\rm m$