Question

If the acceleration of the projective is: a = c s m/s 2 Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s=1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with a velocity of 200 m/s?

Answer 1

c = 4,444.44

Step-by-step explanation:

You have the following expression for the acceleration of the projectile:

`a=cs` (1)

s: distance to the ground of the projectile

To find the value of the constant c you use the following formula:

`v^2=v_o^2+2a \Delta s` (2)

vo: initial velocity = 0 m/s

v: final speed = 200 m/s

Δs: distance traveled by the projectile = 3m - 1.5m = 1.5m

You replace the expression (1) into the expression (2):

`v^2=2(cs)\Delta s`

You do the constant c in the last equation, then you replace the values of v, s and Δs:

`c=(v^2)/(2s\Delta s)=((200m/s)^2)/(2(3m/s^2)(1.5m))=4444.44`