- #1
Hadjiev
- 10
- 0
Here goes:
A sniper shot is fired at over a kilometer away. Furthermore, the target is at the very edge of a building that is 40 m high. Solve for the elevation angle that will be required to score a perfect hit. The initial velocity of the bullet is 854 m/s.
Note: the bullet is not at max height when it hits the target.
I started out by breaking up the sniper into components. The horizontal component came out to 854cos(x) and the vertical 854sin(x).
I solved for the time using the horizontal component; it came out to 1.756/cosx
The time for the horizontal component equals the vertical; thus, I subbed time into d = v1(t) + 1/2(t^2).
I got this: 40 = 854sinx(1.756/cosx) + 4.9(1.756/cosx)^2
This is where I run into a problem. I tried to isolate for x by using various identities and nothing came through. I tried converting everything to tanx, and still no luck. I would greatly appreciate it if someone could give me a few hints. Thanks in advance,
Nicholas
A sniper shot is fired at over a kilometer away. Furthermore, the target is at the very edge of a building that is 40 m high. Solve for the elevation angle that will be required to score a perfect hit. The initial velocity of the bullet is 854 m/s.
Note: the bullet is not at max height when it hits the target.
I started out by breaking up the sniper into components. The horizontal component came out to 854cos(x) and the vertical 854sin(x).
I solved for the time using the horizontal component; it came out to 1.756/cosx
The time for the horizontal component equals the vertical; thus, I subbed time into d = v1(t) + 1/2(t^2).
I got this: 40 = 854sinx(1.756/cosx) + 4.9(1.756/cosx)^2
This is where I run into a problem. I tried to isolate for x by using various identities and nothing came through. I tried converting everything to tanx, and still no luck. I would greatly appreciate it if someone could give me a few hints. Thanks in advance,
Nicholas