Projectile Motion (vectors, max height)?

Okay, here's the question. I've got the first two parts done (if you want to check my answers to see if I'm right, that would be appreciated, but not my main concern), but I'm having trouble on the third part.





A bullet is fired from a gun at ground level at 300 m/s. It hits the ground 3.0 seconds later.





a)At what angle above the horizon was the bullet fired?


(I got a.87 degrees for this part)


b)What was the range (change in position) of the bullet?


(I got 450 meters)


c) What was the maximum height reached by the bullet?





I don't why, I just can't come up with the equation and I've completely blanked on how to do the problem.





Help would be so much appreciated as soon as possible! Thanks so much.

Projectile Motion (vectors, max height)?
to solve the third part, knowing the range and and the angle of elevation, the formulas you are going to use are


1. d= (vfy^2 - viy^2)/2g


d=maximum height


vfy=final vertical velocity= y component of final velocity


vfy = 0, true only at highest point


viy =initial vertical velocity = y component of initial velocity


viy= vi*sin 87


=299.6m/s





g = -9.8m/s^2, negative coz bullet moves upward





d= (0 - 299.6^2)/- 19.6


= 4579.6 meter





or 2. d = viy*t + 0.5*g*t^2


t = time = 3/2 = 1.5 sec, true only at highest point


=299.6*1.5 - 0.5*9.8*1.5^2


=448.5m - 11.03m


=437.5 m





since two answers are different, so your angle is wrong


to solve for the angle use this formula


R = vi^2sin(2*angle)/g = vi*cos(angle) *( t)
Reply:http://hyperphysics.phy-astr.gsu.edu/hba...