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.

113.58 m

Explanation:

Range = Maximum Height

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

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2