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 = 53.0o (as shown), the crew fires the shell at a muzzle velocity of 169 feet per second. How far down the hill does the shell strike if the hill subtends an angle = 41.0o 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?

* The mortar is like a small cannon that launches shells at steep angles.

Explanation / Answer

consider the motion perpendicular to incline

Voy = speed of launch = Vo Sin( + ) = 169 Sin(53 + 41) = 168.6 ft/sec

a = acceleration = - g Cos = - 32.2 Cos41 = - 24.3 ft/s2

t = time of travel

Y = displacement = 0

using the equation

Y = Voy t + (0.5) a t2

0 = 168.6 t + (0.5) (-24.3) t2

t = 13.88 sec

consider the motion parallel to incline :

Vox = speed of launch = Vo Cos( + ) = 169 Cos(53 + 41) = - 11.8 ft/sec

a = acceleration = g Sin = 32.2 Sin41 = 21.1 ft/s2

t = time of travel =13.88

X = displacement = ?

using the equation

X = Vox t + (0.5) a t2

X = (- 11.8) (13.88) + (0.5) (21.1) (13.88)2

X = 1868.72 ft

along the X-direction

Vfx = Vox + a t = -11.8 + (21.1) (13.88) = 281.1 ft/s

along the Y-direction

Vfy = Voy + a t = 168.6 + (- 24.3) (13.88) = - 168.6 ft/s

net speed is given as

Vf = sqrt(Vfx2 + Vfy2) = sqrt((281.1)2 + (168.6)2) = 327.8 ft/s

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