a bullet is fired vertically upwards wth a initial velocity of 300m/s as it leaves the barrel.the rifle barrel is level with the ground and the rifleman stands in a trench.calculate the velocity of the bullet when it is 200m from ground level during the downward movement.
To calculate the velocity of the bullet when it reaches a height of 200m during its downward movement, we can use the equations of motion.
1. First, we need to determine the maximum height reached by the bullet. Since the bullet is fired vertically upwards with an initial velocity of 300 m/s, we know that the final velocity at the highest point will be zero (as the bullet momentarily comes to a stop before falling back down).
Using the equation: v^2 = u^2 + 2as, where v = final velocity, u = initial velocity, a = acceleration, and s = displacement.
Replacing v = 0, u = 300 m/s, and a = -9.8 m/s^2 (acceleration due to gravity), we can solve for the displacement s:
0^2 = (300 m/s)^2 + 2(-9.8 m/s^2)s
Simplifying the equation gives us:
0 = 90000 - 19.6s
19.6s = 90000
s = 90000 / 19.6
s ≈ 4591.8 m
Therefore, the maximum height reached by the bullet is approximately 4591.8m.
2. Next, we can calculate the time it takes for the bullet to fall from its maximum height to a height of 200m. Since the bullet is in free fall, it undergoes uniformly accelerated motion downward with an acceleration of -9.8 m/s^2.
Using the equation: s = ut + (1/2)at^2, where s = displacement, u = initial velocity (zero in this case), a = acceleration (-9.8 m/s^2), and t = time, we can solve for t:
200m = 0 + (1/2)(-9.8 m/s^2)t^2
Simplifying the equation gives us:
200m = (-4.9 m/s^2)t^2
t^2 = -40.8163
t ≈ √(-40.8163) (ignoring the negative square root as it doesn't make physical sense)
Therefore, the time taken for the bullet to fall from its maximum height to a height of 200m is approximately √40.8163 seconds.
3. Finally, we can determine the velocity of the bullet when it is 200m from the ground level during its downward movement. We can use the equation: v = u + at, where v = final velocity, u = initial velocity, a = acceleration (-9.8 m/s^2), and t = time.
Replacing u = 0, a = -9.8 m/s^2, and t = √40.8163 s, we can solve for v:
v = 0 + (-9.8 m/s^2)(√40.8163 s)
v ≈ -9.8 m/s^2 * 6.39013 s
v ≈ -63.73 m/s
Therefore, the velocity of the bullet when it is 200m from the ground level during its downward movement is approximately -63.73 m/s (negative sign indicating downward direction).
To determine the downward velocity of the bullet when it is 200m from ground level, we can use the principles of projectile motion.
Firstly, let's understand the key details given in the problem:
- The bullet is fired vertically upwards from the barrel of a rifle.
- The initial velocity of the bullet is 300m/s.
- The rifle barrel is level with the ground.
- The rifleman stands in a trench.
Now, since the bullet is fired vertically upwards, it will reach its maximum height and then start falling downward due to gravity. At the maximum height, the bullet's velocity will be zero.
To find the time it takes for the bullet to reach its maximum height, we can use the equation:
v = u + at,
where:
v = final velocity (0 m/s at the maximum height)
u = initial velocity (300 m/s)
a = acceleration (acceleration due to gravity, which is approximately -9.8 m/s^2)
t = time
Solving for t, we have:
0 = 300 - 9.8t
9.8t = 300
t = 300 / 9.8 ≈ 30.6 seconds
Therefore, it takes approximately 30.6 seconds for the bullet to reach its maximum height.
Now, to find the velocity of the bullet when it is 200m from ground level during the downward movement, we can use the equation:
v = u + at,
where:
v = final velocity (what we want to find)
u = initial velocity (0 m/s at the maximum height)
a = acceleration (acceleration due to gravity, which is approximately -9.8 m/s^2)
t = time
We need to find the time it takes for the bullet to fall 200m from its maximum height. Since the time to reach the maximum height is 30.6 seconds, we can use symmetry to determine the time for the downward movement. Thus, it will be half the total time for the complete upward and downward motion.
t = 30.6 / 2 = 15.3 seconds
Now we can substitute the values into the equation:
v = 0 + (-9.8) * 15.3
v ≈ -149.94 m/s
Therefore, the velocity of the bullet when it is 200m from ground level during the downward movement is approximately -149.94 m/s. The negative sign indicates that the velocity is directed downwards.