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1.a machine gun fires a ball with an initial velocity of 600m/s with an elevation of 30° with respect to the ground neglecting air resistance calculate:
a.the maximum height that can be reached?
b.the time of flight of the bullet?
c.the maximum horizontal displacement of the ired bullet?​

Respuesta :

pstnonsonjoku

Answer:

See explanation

Explanation:

a) maximum height of a projectile = u sin^2θ/2g

H= 600 × (sin 30)^2/2 × 10

H= 7.5 m

b) Time of flight

t= 2u sinθ/g

t= 2 × 600 sin 30/10

t= 60 seconds

Range

R= u^2sin2θ/g

R= (600)^2 × sin2(30)/10

R= 31.2 m

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