Asked by ally
The initial kinetic energy imparted to a
0.72 kg bullet is 1777 J. The acceleration of gravity is 9.81m/s2.Neglecting air resistance, find the range of this projectile when it is fired at an angle such that the range equals the maximum height attained.
Answer in units of km.
0.72 kg bullet is 1777 J. The acceleration of gravity is 9.81m/s2.Neglecting air resistance, find the range of this projectile when it is fired at an angle such that the range equals the maximum height attained.
Answer in units of km.
Answers
Answered by
drwls
First calculate the initial velocity Vo, using
(M/2)Vo^2 = 1777 J
If the range equals the maximum height,
2(Vo^2/g)sinA*cosA = (1/2)(Vo^2/g)sin^2A
2 cosA = (1/2)sin A
tanA = 4 A = 76.0 degrees
Use that angle in either the range or the max height formula
(M/2)Vo^2 = 1777 J
If the range equals the maximum height,
2(Vo^2/g)sinA*cosA = (1/2)(Vo^2/g)sin^2A
2 cosA = (1/2)sin A
tanA = 4 A = 76.0 degrees
Use that angle in either the range or the max height formula
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