A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.

Question

3 years ago

10

A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.

Enemy forces are charging up the hill and it is necessary for the crew to spring into action. Angling the mortar at an angle of
θ
=
49.0
∘
(as shown), the crew fires the shell at a muzzle velocity of
156
ft/s.

How far
d
down the hill does the shell strike if the hill subtends an angle
ϕ
=
41.0
∘
from the horizontal? Ignore air friction.

A coordinate system is centered on a point at the top of a hill. A velocity vector of magnitude v subscript zero extends from the origin of the coordinate system and makes a counterclockwise angle theta with the positive horizontal axis. The hill slopes downward from the origin, and the slope makes a clockwise angle phi with the positive horizontal axis. A parabolic trajectory shows that the shell lands a distance d downhill, measured along the length of the slope.
d
=

m

How long will the mortar shell remain in the air?
time:

s

How fast will the shell be traveling when it hits the ground?
speed:

m
/
s

Physics

1 answer:

Elena-2011 [213] · Answer 1 · 2022-05-25T20:46:21+03:00

1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

1)

The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

$y=(u sin \theta)t-\frac{1}{2}gt^2$ (1)

where:

$u sin \theta$ is the initial vertical velocity of the shell, with $u=156 ft/s$ and $\theta=49.0^{\circ}$

$g=32 ft/s^2$ is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

$x=(ucos \theta)t$

where $u cos \theta$ is the initial horizontal velocity of the shell.

We can re-write this last equation as

$t=\frac{x}{u cos \theta}$ (1b)

And substituting into (1),

$y=xtan\theta -\frac{1}{2}gt^2$ (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of $\alpha=-41.0^{\circ}$ from the horizontal, so we can write the position (x,y) of the hill as

$y=x tan \alpha$ (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

$xtan\alpha = xtan \theta - \frac{1}{2}gt^2$

Substituting (1b) into this equation,

$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$