A shell is ﬁred from the ground with an initial speed of 1.70 x 10^3 m/s at an initial angle of 55.0° to the horizontal, Neglecting air resistance, what is the shell's horizontal range and the amount of time the shell is in motion?

Nov 13, 2015

$\left(a\right)$ $277.1 \text{km}$

$\left(b\right)$ $284.2 \text{s}$

Explanation:

$\left(a\right)$ The range, assuming level ground, is given by:

$d = \frac{{v}^{2}}{g} \sin \left(2 \theta\right)$

$\therefore d = \frac{{\left(1.7 \times {10}^{3}\right)}^{2} \times 0.9397}{9.8}$

$d = 0.2771 \times {10}^{6} \text{m}$

$d = 277.1 \text{km}$

$\left(b\right)$

The horizontal component of velocity is constant and is equal to $v \cos \theta$.

So $t = \frac{s}{v} = \frac{0.2771 \times {10}^{6}}{1.7 \times {10}^{3} \times \cos 55}$

$\therefore t = \frac{277.1}{1.7 \times 0.573}$

$t = 284.2 \text{s}$