physics problem involving projectile motion and force?
a 32g bullet is fired from a rifle at 380m/s into a water bottle at pointblank range the bullet travels completely through the 12cm long bottle located 2m above the ground it finally comes to earth at a point 3m downrange from the point where it left the water find the average force exerted by the water on the bullet
i found the total time it took for the bullet to hit the ground by using the y-dir and found the total time it would take had it not gone through the bottle with the x-dir and subtracted but i don't get the right answer. anyone know how to solve this?
- Anonymous1 decade agoFavorite Answer
F be the force exerted by the water,
m be the mass of the bullet,
a be its acceleration as it goes through the water,
u be its entry speed,
v be its exit speed,
s be the length of the water bottle,
x, y be the horizontal and vertical distances travelled after leaving the bottle,
t be the time of flight after leaving the bottle,
g be the acceleration due to gravity.
Passing through the bottle:
F = - ma
The loss of KE is:
Fs = m(u^2 - v^2) / 2 ...(1)
After leaving the bottle:
x = vt
y = gt^2 / 2
= gx^2 / (2v^2)
v^2 = gx^2 / (2y) ...(2)
Eliminating v^2 from (1) and (2):
Fs = m(u^2 - gx^2 / [ 2y ])
F = m(u^2 - gx^2 / [ 2y ]) / (2s)
= 0.032(380^2 - 9.81 * 3^2 / [ 2 * 2 ]) / (2 * 0.12)
= 1.93 * 10^4 N.
- meluskyLv 45 years ago
I wish no one makes use of this reply, when you consider that it is horribly mistaken. The answerer forgot to take acceleration under consideration while settling on whether or not or no longer the projectile would transparent the mountain top. At a forty five measure perspective, the shell fired from the enemy send could simplest make it to 1520 meters prime while out 2500m in which the mountain top is.