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

Question

A)  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 ? = 66.0o (as shown), the crew fires the shell at a muzzle velocity of 162 feet per second. How far down the hill does the shell strike if the hill subtends an angle ? = 36.0o from the horizontal? (Ignore air friction.)

B)  How long will the mortar shell remain in the air?

C)  How fast will the shell be traveling when it hits the ground?

Explanation / Answer

To make this problem simpler, lets convert the velocity to m/s.
1 ft * 12 in/ft * 2.54 cm/in * 1 m/100 cm = 0.3048 m
Initial velocity = 162 * 0.3048 = 49.3776 m/s
I will assume that the initial angle is 66? above horizontal.
Initial vertical velocity = 49.3776 * sin 66
Initial horizontal velocity = 49.3776 * cos 66

The horizontal component of the initial velocity is constant. The vertical component is decreasing at the rate of 9.8 m/s each second.

Since the final angle is 36? above horizontal, tan 36 = final vertical velocity final horizontal velocity.

tan 36 = final vertical velocity 49.3776 * cos 66
Final vertical velocity = 49.3776 * cos 66 * tan 36

Now we know the final and initial vertical velocities. Lets use the following equation to determine the vertical distance the shell dropped.

vf^2 = vi^ 2 + 2 * a * d, a = -9.8 m/s^2

(49.3776 * cos 66 * tan 36)^2 = (49.3776 * sin 66)^2 + 2 * -9.8 * d
(49.3776 * cos 66 * tan 36)^2 (49.3776 * sin 66)^2 = -19.6 * d
d = [(49.3776 * cos 66 * tan 36)^2 (49.3776 * sin 66)^2] -19.6
212.897281 2034.731664 = -1821.834383
d = -1821.834383 -19.6
92.95 meters
This is the vertical distance the shell dropped.

Distance down the hill = vertical distance sin 36
Distance down the hill = 92.95 sin 36= 158.137 meters
Lets convert to feet by dividing by 0.3048
Distance down the hill = (215.1102522 sin 41) 0.3048 = 518.822 ft

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