A bullet is shot from a height of 5 meters with a velocity of 230 m/s. How long before the bullet hits the ground? How far will it travel?

Question

Free Fall Kinematic Equations which may help with this problem.

Vf=Vi + 9.8*t

 

d=Vi*t + .5*9.8*t²

 

Vf² = Vi² + 2*9.8*d

 

d= [Vi + Vf]*t/2

Answer 1

A bullet is shot from a height of 5 meters with a velocity of 230 m/s. How long before the bullet hits the ground? How far will it travel?

This question is incomplete because it does not specify the direction in which the bullet was fired.

Based on some of the results expected, It appears that the direction is horizontal.

Every bullet, once fired, becomes a projectile. This means that as soon as it leaves the gun, the only force acting on it is the force of gravity pulling it down. The bullet will be in free fall.

V_i = 0

g = 9.81m/s²

d = 5m

t = ?

Since we do not have a value for V_f, the only suitable equation is the second one. I'll let you finish it yourself.

Projectile motion is a rather curious type of motion and most students have difficulty recognizing the finer details. Once you understand those, you will find projectile motion quite straight forward.

For the time (t), remember there is no force in the horizontal direction so no acceleration. This means the velocity 230m/s is uniform. Use the expression V = d/t to find t.

I hope this helps. Let me know if you need further clarification.

Mr. A

Answer 2

Because the bullet is in free fall it will accelerate at 9.81 m/s^2

using d = .5at^2 we can solve for t since we have a (9.81m/s^2) and d is 5 meters

so t = sqrt(2*5/9.81) = 1.009 seconds

so now we have the time. With the time and the velocity we can find the distance:

velocity = distance/time becomes 230 = distance/1.009 = 232 m/s

hope this helps!