A projectile motion is fired horizontally from a gun that is 45.0m above the flat ground,emerging from the gun with a speed of 250m/s.What is the magnitude of the vertical component of its velocity as it strikes the ground?

1 Answer
Hugo C.
Oct 12, 2015

#v = 29.71 m/s#

Explanation:

I have taken air resistance as negligible, as it is extremely difficult to calculate velocities with it, and the aerodynamics of the projectile would be needed. As we are only asked for the velocity in the y direction, we can ignore the initial horizontal velocity, #u = 250m/s#, and assume that:
Initial velocity: #u = 0 m/s#
Acceleration (gravity): #a = -9.81 m/(s^2)#
Displacement: #s = 45m#

Therefore, we can use the formula: #v^2 = u^2 + 2as#.

#v^2 = 0^2 + 2 * -9.81 * 45#,
#v^2 = 0 + 882.9#,
#v^2 = 882.9#,
#v = sqrt(882.9)#,
#v = 29.71 m/s#.

Related questions
Impact of this question
12653 views around the world
You can reuse this answer
Creative Commons License