height of projectile is maximum at an angle?
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I believe that many people misinterpreted the question. The question addresses and asked, "At which angle, the height of projectile is maximum."
The question seems to address the situation where the projectile is fired at an initial inclination angle A with a muzzle velocity v in gravity a. Thus the projectile will leave the muzzle at an angle and A rise up and away from the original position and then drops down again. Now a short little man looking and following the projectile, will raise his head and then lower it. We are required to find what is the maximum angle B this short little man looks up as he follows the projectile.
The question asks what is the angle subtended from the original position when the height is maximum, for any value of firing angle A and velocity v.
Well I make that the Tangent of the angle B equals half the tangent of angle A.
Time of flight to max height equals...................(v sinA)/a
Y=maximum height reached .............................(v sinA)(v sinA)/2a
X=half the range to maximum height ...............(v cosA)(v sinA)/a
Tangent of B ..................... equals Y/X = ( tangent A)/2
for maximum height B= 90 degrees.
Interesting how different people understand the same question in a different manner so the person who asks the question must be so careful.
The question seems to address the situation where the projectile is fired at an initial inclination angle A with a muzzle velocity v in gravity a. Thus the projectile will leave the muzzle at an angle and A rise up and away from the original position and then drops down again. Now a short little man looking and following the projectile, will raise his head and then lower it. We are required to find what is the maximum angle B this short little man looks up as he follows the projectile.
The question asks what is the angle subtended from the original position when the height is maximum, for any value of firing angle A and velocity v.
Well I make that the Tangent of the angle B equals half the tangent of angle A.
Time of flight to max height equals...................(v sinA)/a
Y=maximum height reached .............................(v sinA)(v sinA)/2a
X=half the range to maximum height ...............(v cosA)(v sinA)/a
Tangent of B ..................... equals Y/X = ( tangent A)/2
for maximum height B= 90 degrees.
Interesting how different people understand the same question in a different manner so the person who asks the question must be so careful.
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190 is the highest of projectile is the maximum angle
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