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:
[tex]y=(u sin \theta)t-\frac{1}{2}gt^2[/tex] (1)
where:
[tex]u sin \theta[/tex] is the initial vertical velocity of the shell, with [tex]u=156 ft/s[/tex] and [tex]\theta=49.0^{\circ}[/tex]
[tex]g=32 ft/s^2[/tex] 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
[tex]x=(ucos \theta)t[/tex]
where [tex]u cos \theta[/tex] is the initial horizontal velocity of the shell.
We can re-write this last equation as
[tex]t=\frac{x}{u cos \theta}[/tex] (1b)
And substituting into (1),
[tex]y=xtan\theta -\frac{1}{2}gt^2[/tex] (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 [tex]\alpha=-41.0^{\circ}[/tex] from the horizontal, so we can write the position (x,y) of the hill as
[tex]y=x tan \alpha[/tex] (3)
Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):
[tex]xtan\alpha = xtan \theta - \frac{1}{2}gt^2[/tex]
Substituting (1b) into this equation,
[tex]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[/tex]
Which has 2 solutions:
x = 0 (origin)
and
[tex]tan \theta - tan \alpha=\frac{gx}{2u^2 cos^2 \theta}=0\\x=(tan \theta - tan \alpha) \frac{2u^2 cos^2\theta}{g}=1322 ft[/tex]
So, the distance d down the hill at which the shell strikes the hill is
[tex]d=\frac{x}{cos \alpha}=\frac{1322}{cos(-41.0^{\circ})}=1752 ft=534 m[/tex]
2)
In order to find how long the mortar shell remain in the air, we can use the equation:
[tex]t=\frac{x}{u cos \theta}[/tex]
where:
x = 1322 ft is the final position of the shell when it strikes the hill
[tex]u=156 ft/s[/tex] is the initial velocity of the shell
[tex]\theta=49.0^{\circ}[/tex] is the angle of projection of the shell
Substituting these values into the equation, we find the time of flight of the shell:
[tex]t=\frac{1322}{(156)(cos 49^{\circ})}=12.9 s[/tex]
3)
In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.
The horizontal component of the velocity is constant and it is
[tex]v_x = u cos \theta =(156)(cos 49^{\circ})=102.3 ft/s[/tex]
Instead, the vertical component of the velocity is given by
[tex]v_y=usin \theta -gt[/tex]
And substituting at t = 12.9 s (time at which the shell strikes the hill),
[tex]v_y=(156)(cos 49^{\circ})-(32)(12.9)=-310.4ft/s[/tex]
Therefore, the final speed of the shell is:
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{(102.3)^2+(-310.4)^2}=326.8 ft/s=99.6 m/s[/tex]
Learn more about projectile motion:
brainly.com/question/8751410
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