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 0 = 65.0° (as shown), the crew fires the shell at a muzzle velocity of 148 ft/s. How far d down the hill does the shell strike if the hill subtends an angle p = 37.0° from the horizontal? Ignore air friction. d = m How long will the mortar shell remain in the air? time: How fast will the shell be traveling when it hits the ground? speed: m/s
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 0 = 65.0° (as shown), the crew fires the shell at a muzzle velocity of 148 ft/s. How far d down the hill does the shell strike if the hill subtends an angle p = 37.0° from the horizontal? Ignore air friction. d = m How long will the mortar shell remain in the air? time: How fast will the shell be traveling when it hits the ground? speed: m/s
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Step 1
Consider the following figure which represents the motion of the shell.
Here, y represents the vertical displacement of the shell and x represents the horizontal displacement of the shell.
According to geometry, the values of x and y are given in the following way.
The vertical and horizontal components of the initial velocity and acceleration are given in the following way.
Here, g is the acceleration due to gravity.
In this case,
Let the time for which the shell remains in the air be t.
According to Newton’s equation of motion, the value of d in terms of t and v0 is calculated in the following way.
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