Answer to Question #229662 in Physics for April

Question

A bullet is fired at an angle of 30° above the horizontal with a velocity of 600m/s. Find the range the time it returns to the same level and the position and velocity of the bullet 40 s after after it is fired

Answer 1

Given:

$v_0=600\:\rm m/s$

$\theta=30^{\circ}$

$t_1=40\:\rm s$

The range

$R=\frac{v_0^2\sin2\theta}{g}=\frac{600^2\sin 60^{\circ}}{9.8}=31813\:\rm m$

The time of motion

$t=\frac{2v_0\sin\theta}{g}=\frac{2*600\sin 30^{\circ}}{9.8}=61\:\rm s$

After 40 s

$x=v_0\cos\theta *t_1=600\cos 30^{\circ}*40=20784\:\rm m$

$y=v_0\sin\theta *t_1-\frac{gt_1^2}{2}$

$y=600\sin 30^{\circ}*40-\frac{9.8*40^2}{2}=4160\:\rm m$

The speed of a bullet

$v=\sqrt{(v_0\cos\theta)^2+(v_0\sin\theta-gt_1)}$

$v=\sqrt{(600\cos 30^{\circ})^2+(600\sin30^{\circ}-9.8*40)^2}\\=528\:\rm m/s$