A mortar* crew is positioned near the top of a steep hill. Enemy forces are char

Question

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 ? = 49.0o (as shown), the crew fires the shell at a muzzle velocity of 202 feet per second. How far down the hill does the shell strike if the hill subtends an angle ? = 35.0o from the horizontal? (Ignore air friction.) How long will the mortar shell remain in the air? How fast will the shell be traveling when it hits the ground?

Explanation / Answer

Vyi = 209sin65 = 189ft/s
Vxi = 209cos65 = 88ft/s
h(t) = Vyi*t - 16*t^2 = 189*t - 16*t^2
Let y(t) be the surface of the hill as a function of time
y(t) = -Vxi*t*tan39 = -71.5*t
If we set h(t)=y(t) we can find when the height of the mortar is the same as the surface of the hill

189t -16t^2 = -71.5t => t = [189+71.5]/16 = 16.3s
At that moment the mortar will be -71.5*16.3/sin39 = 1853.6 ft down the hill
Which means you would have to walk 1853.6 ft from the launch point down the hill to reach the point of impact.
The vertical distance down is 71.5* 16.3 = 1166.5ft vertically beneath the launch point
The horizontal distance out would be 1440.5 ft from the launch point

OR YOU CAN SAY

y = h + x·tan - g·x² / (2v²·cos²)
y = x * sin(-35º)
h = 0
x = ?
= 49º
v = 202 ft/s
x * sin(-35) = 0 + xtan49 - 32.2x² / (2*202²*cos²49)
-0.573x = 1.15x - 0.185x²
0 = 1.723x - 0.185x²
x = 0 ft, 9.313 ft

Along the slope, it's 9.313ft/cos(-35º) = 11.37 ft
y = 11.37 * sin(-49) = -15.06 ft
Time at/above launch height = 2·Vo·sin/g = 2 * 202ft/s * sin49 / 32.2 ft/s² = 9.8 s
initial vertical velocity Vv = 202ft/s * sin49º = 157.7 ft/s
so upon returning to launch height, Vv = -157.7 and time to reach the ground is
-15.06 ft = -157.7 * t - ½ * 32.2ft/s² * t²
0 = 15.06 - 157.7t - 16.1t²
solve t
To the total time of flight is 9.8s + t

at impact, Vv = Vvo * at = -157.7ft/s - 32.2ft/s² * 5.3s = -328 ft/s
Vx = 202ft/s * cos49º = 132.52 ft/s
V = ((Vx)² + (Vy)²) = 353.76 ft/s

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