A projectile is fired horizontally from a gun that is 44.0 m above flat ground,

Question

A projectile is fired horizontally from a gun that is 44.0 m above flat ground, emerging from the gun with a speed of 250 m/s. How long does the projectile remain in the air? At what horizontal distance from the firing point does it strike the ground? What is the magnitude of the vertical component of its velocity as it strikes the ground? First decide: What is the initial vertical velocity? Then find how long the projectile takes to fall the given height as if the motion is one-dimensional. During that falling time, the projectile is moving horizontally. First decide: What is the horizontal acceleration? Then find the horizontal distance. Find the vertical velocity component at the end of the fall as if the motion is one-dimensional.

Explanation / Answer

Projectile motion

a] Use the kinematic equation along the vertical direction

y= Viy* t+1/2gt2=0+4.9 t2    [ Viy(Initial vertical velocity)= 0 m/s]

44=4.9 t2

t= 3sec

b] Use the kinematic equation along the horizontal direction

Here horizontally velocity remains constant and acceleration along the horizontal direction =0 m/s2

Vx= 250 m/s---constant

ax=0 m/s2

Horizontal distance X= Vx* t= 250 m/s* 3s= 750 m

c] Vertically

Vfy= Viy+gt= 0+(9.8* 3) = 29.4 m/s    [ Viy(Initial vertical velocity)= 0 m/s]

The magnitude of the vertical component of its velocity as it strikes the ground is 29.4 m/s

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