a 25.0 kg shell is fired out to sea from a navy gun that is situated at sea leve

Question

a 25.0 kg shell is fired out to sea from a navy gun that is situated at sea level. the initial speed of the shell is 200m/s and it is fired at an angle of 30 degrees above horizontal.

a) ignoring air resistance, what is the maximum height above sea level that the shell reaches?

b) what is the velocity (magnitude and direction) of the shell when it hits the surface of the sea?

NB this question can be answered using newton's laws and calculating accelerations. No marks will be given for answering this way.

Explanation / Answer

a) To do this, break the trajectory into vectors. Only the vertical vector is necessary. To find the initial vertical velocity, find (200m/s)sin(30) = 100m/s The max height will be when the vertical velocity = 0. (Velocity @ Max ) = (Initial Velocity ) + (Acceleration due to gravity )(Time) Solve for time to find how long it takes to reach this maximum. This will give you t=10 seconds. Plug this value into the distance formula, understanding that the initial height is at sea level (therefore zero). h=vt + (1/2)at^2 = (100m/s)(10s)+(1/2)(-10m/s^2)(10s)^2 = 500 m This gives a maximum height of 500 meters. b) By definition of projectile motion (which this is, the shell is the projectile), if the path begins at an angle upward starting with a height of zero, the final velocity will be the same as the initial, but with the angle in the opposite direction. Therefore, the shell will be moving with a velocity of 200 m/s toward the ground with an angle of 30 degrees. You could also put this in vector form, where the horizontal component is (200)cos(30) and the vertical is -(200)sin(30) (it's negative because it's now moving downward). This will give you 178.2 m/s horizontally and -100 m/s vertically.

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