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The initial kinetic energy imparted to a 0.97 kg bullet is 1150 J. The acceleration of gravity is 9.81 m/s 2 . 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.

Respuesta :

Answer:

113.58 m

Explanation:

Range = Maximum Height

[tex]\frac{u^{2}Sin2\theta }{g}=\frac{u^{2}Sin^{2}\theta }{2g}[/tex]

tan θ = 4

θ = 76 degree

Let u be the velocity of projection

K = 1/2 x m x u^2

1150 = 0.5 x 0.97 x u^2

u = 48.7 m/s

Range = [tex]\frac{u^{2}Sin2\theta }{g}

R = (48.7 x 48.7 x 2 x Sin 76 x Cos 76) / 9.8 = 113.58 m

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