A projectile is projected from the origin with a velocity of 45.0 m/s at an angle of 30.0 degrees above the horizontal. What is the velocity when the projectile hits the ground?

velocity = 44.98 m/s

Explanation:

given data

velocity = 45.0 m/s

angle = 30.0 degrees

solution

first we get here time that is

time = 2u × sin θ ÷ g   .............1

time = 2×45 × sin 30 ÷ 9.8

time = 4.59 sec

and

velocity =   .....................2

here

vx = 45 cos30 = 38.97 m/s

vy = uy - gt

vy = 45 sin30 - 9.8 ×  4.59  = -22.48 m/s

so velocity will be

velocity =

velocity = 44.98 m/s

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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