physics: Velocity and acceleration.

[shawnarae]

Help it is the end of the quarter and this stuff is hard.....Here is my question.....step by step would be great...okay here goes
155MM hOWITZER COASTAL ARTILLERY GUN IS FIRED OFF AT A TRAJECTOIRY ANGLE OF 35 DEGREES OFF THE HORIZON. iF THE MUZZLE VELOCITY IS 827M/S WHAT IS THE MAXIMUM ELEVATION OF THE PROJECTILE DURING FLIGHT TIME? AND HOW FAR DOWN RANGE WILL THE PROJECTILE TRAVEL? aSSUME TEMP,WIND AND HUMIDITY ARE NEGLIGABLE
Please help I am really struggling...math is not my gig....thank you so much :? :wink:

 

[tkhunny]

You'll need the sine and cosine of 35º. The sine will help you with height and time. The cosine will help you with range.

 

[TchrWill]

Help it is the end of the quarter and this stuff is hard.....Here is my question.....step by step would be great...okay here goes
155MM hOWITZER COASTAL ARTILLERY GUN IS FIRED OFF AT A TRAJECTOIRY ANGLE OF 35 DEGREES OFF THE HORIZON. iF THE MUZZLE VELOCITY IS 827M/S WHAT IS THE MAXIMUM ELEVATION OF THE PROJECTILE DURING FLIGHT TIME? AND HOW FAR DOWN RANGE WILL THE PROJECTILE TRAVEL? aSSUME TEMP,WIND AND HUMIDITY ARE NEGLIGABLE.

The height reached derives from h = Vo^2(sin^2(µ))/2g.

The distance traveled derives from d = Vo^2(sin(2µ))/g

h = the maximum height in meters
d = the horizontal distance traveled in meters
Vo = the initial velocity in meters/second
g = the acceleration due to gravity = 9.8M/S^2
µ = the angle of elevation of the gun from the horizontal in degrees

I'll leave the hard part for you.