A projectile is fired horizontally from a gun that is 59.0 m above flat ground,

Question

A projectile is fired horizontally from a gun that is 59.0 m above flat ground, emerging from the gun with a speed of 210 m/s. (a) How long does the projectile remain in the air? (b) At what horizontal distance from the firing point does it strike the ground? (c) What is the magnitude of the vertical component of its velocity as it strikes the ground? A projectile is fired horizontally from a gun that is 59.0 m above flat ground, emerging from the gun with a speed of 210 m/s. (a) How long does the projectile remain in the air? (b) At what horizontal distance from the firing point does it strike the ground? (c) What is the magnitude of the vertical component of its velocity as it strikes the ground?

Explanation / Answer

a) The vertical distance travelled by the projectile before it strikes the ground is

y = (uy ) t - (1/2) g t2

uy = Vertical component of intial speed of the projectile = 0 m/s ( Projectile fired horizontally )

t = Time of flight

g = Acceleration due to gravity =9.8 m/s2.

y = Vertical distance from the ground to gun = - 59.0 m

On substituting values we get

- 59.0 m = - (1/2) (9.8 m/s2 ) t2

t =3.47 s

Therefore the projectile remains in the air for 3.47 s.

b)The horizontal distance travelled by the projectile before it strikes the ground is

x = ux t

here x = horizontal distance travelled by the projectile before it strikes the ground

ux = Horizontal speed of the projectile = 210 m/s

t = Time of flight of the projectile =3.47 s

On substituting all values we get

x = (210 m/s) (3.47 s) = 728.70 m

Therefore the horizontal distance travelled by the projectile before it strikes the ground is 728.70 m .

c)The magnitude of the vertical component of its velocity as it strikes the ground is

|vy| =| -g t |

{g =9.8 m/s2  , t =3.47 s }

|vy| = (9.8 m/s2) (3.47 s)

|vy| = 34.01 m/s

Therefore the magnitude of the vertical component of its velocity as it strikes the ground is 34.01 m/s.

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