A bullet was fired with initial velocity of 966 ft/s directly upward.
conditions:
a. What is its acceleration 15.0 after it was fired?
b. What is it's velocity at the peak of its fight?
c. What is its acceleration at the peak of it's fight?
FORMULA ONLY :DD THANK YOU IN ADVANCE!

1

Answers

2014-08-12T20:53:53+08:00


first Identify the given components:
-delta Y
-delta X
-delta Time
-gravity
-initial velocity on both X and Y axis

Use the following formulas w/c would best fit :

for X component                     for Y component

Xf = X1 + Vxi(t)                   Yf = Yi+ Vyi(t) - 1/2 (g)(timesquared)

Vectors:

Vxi = Vi cos theta
Vyi = Vi sin theta

Note: if possible, don't limit yourself in one equation. Recall every equation you know. I Hope I could help :D


                                         


0
A. It's acceleration will not change, it is still -9.8 m/s^2. B. The velocity at the peak is 0 m/s. No need to solve. C. The acceleration is constant -9.8 m/s^2. It does not change. :] Is this right?