A projectile is fired horizontally from a gun that is 45.0 mabove flat ground, emerging from the gun with a speed of250 m/s. (a) How long does the projectile remain in the air?(b) At what horizontal distance from the firing point does itstrike the ground? (c) What is the magnitude of the verticalcomponent of its velocity as it strikes the ground?

Sure, let's solve this step by step:

(a) How long does the projectile remain in the air?

We can use the equation of motion to find the time it takes for the projectile to hit the ground. The equation is:

h = 0.5 * g * t^2

where: h = height = 45.0 m g = acceleration due to gravity = 9.8 m/s^2 t = time

Rearranging the equation to solve for t gives:

t = sqrt((2h)/g) = sqrt((245.0)/9.8) = 3.03 s

So, the projectile remains in the air for approximately 3.03 seconds.

(b) At what horizontal distance from the firing point does it strike the ground?

The horizontal distance can be found using the equation:

d = v * t

where: d = distance v = velocity = 250 m/s t = time = 3.03 s

So, d = 250 * 3.03 = 757.5 m

Therefore, the projectile strikes the ground at a horizontal distance of approximately 757.5 m from the firing point.

(c) What is the magnitude of the vertical component of its velocity as it strikes the ground?

The vertical component of the velocity can be found using the equation:

v = g * t

where: v = vertical velocity g = acceleration due to gravity = 9.8 m/s^2 t = time = 3.03 s

So, v = 9.8 * 3.03 = 29.7 m/s

Therefore, the magnitude of the vertical component of its velocity as it strikes the ground is approximately 29.7 m/s.