The initial kinetic energy imparted to a 0.0190 kg bullet is 1497 J. (a) Assumin

Question

The initial kinetic energy imparted to a 0.0190 kg bullet is 1497 J. (a) Assuming it accelerated down a 1.00 m long rifle barrel, estimate the average power delivered to it during the firing. kW (b) 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. km Assume that the firing height is negligible and that the bullet lands at the same elevation from which it was fired. Use constant-acceleration equations in the horizontal and vertical directions to find the range of the bullet and its maximum height. The bullet's initial speed can be determined from its initial kinetic energy.

Explanation / Answer

Power = Energy/ Time ; Kinetic energy = 1/2*m*v2

Substituting the given values, we get v = sqrt(2*KE/m) = 396.96m/s

We know, v - v0 = at ; v2 - V20 = 2as

Here v0 = 0; v = 396.96m/s; s = 1.0m

Substituting above values we get,

396.96 = at------------(1)

396.962 = 2a ------------(2)

From (1) & (2) we get a= 78789.94m/s2 t= 5.04 ms

Therefore average power delivered to the bullet P = 1497/(5.04 x 10-3) = 297.13kW

a) P = 297.13 kW

Maximum height attained by the bullet = H = v2(sin2A)/(2g)

Range of the bullet = R = v2(sin2A)/g

Given R = H i.e. (sin2A)/2 = (sin2A) (sin2A = 2sinAcosA)

Solving we get A = tan-4 ; So A = 75.960

So R = v2(sin2A)/g; v=396.96m/s ; A = 75.960 ; g = 9.8m/s2

so R = 7.57km or 7566.73m

b) R = 7.57 km

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