25,000 Feet = 7620m
PE = mgh where m is mass, g is gravity accel: 9.8 n h is height
= 90 x 9.8 x 7620
= 6720840J
= 6.72MJ
F = ma where m is mass, a is accel = gravity = 9.8
= 90 x 9.8
= 882N
Accel = gravity = 9.8m/s^2
KE = 1/2mv^2 where m is mass n v is vel
if no wind resistance, PE leaving airplane = KE at net
6720840 = 1/2 x 90 x v^2
v^2 = 149352
v = 386.5m/s
Answer:
≅50°
Explanation:
We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:
Δx=V₀t+at²/2
And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:
Δx=(V₀cosθ)t+at²/2
Now luckily we are given everything we need to solve (or you found the info before posting here):
- Δx=760 m
- V₀=87 m/s
- t=13.6 s
- a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!
With that we can plug the values in to get:



