A projectile is fired horizontally with an initial speed of 50.0 m/s. Neglect air resistance. a) What is the magnitude of the displacement of the projectile 3.00 s after it is fired?b) What is the speed of the projectile 3.00 s after it is fired?c) What is the magnitude of the acceleration of the projectile 3.00 s after it is fired?

a) 156 meters

b) 58 m/s

c) 9.8 m/s^2

Explanation:

a) The horizontal displacement X = V *t

X = 50 * 3 = 150 meters

Vertical displacement = y = – g * t² / 2

Y = -9.8 *3*3/2 = -4.9*9 = 44.1

Net Displacement = Sqrt (X^2 +Y^2) = 156.34 = 156 meters

b) Speed of the projectile

vx = 50 m/s

vy = t*g = 9.8*3 = 29.4

V = Sqrt (vx^2 + vy^2) = 58 m/s

c) Acceleration is the acceleration due to gravity = 9.8 m/s^2

explanation:

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answer:

yes

explanation: