A bullet is fired directly upward from a rifle at ground level. Assuming a velocity muzzle of 490 m/s, and neglecting air resistance, find:
A.height of the highest point reached.
B.time required to reach this point.
C.total time before the bullet returns to the ground.
D.speed of the bullet upon hitting the ground.
please..somedaY answer this -_-
mv²/2=mgh
h= v²/2g=....
v(fin)=v-gt
v(fin)=0
t=v/g=.....
t(total) =2t
v(ground)=v=490 m/s
A. The height of the highest point reached can be determined using the kinematic equation. Assuming the acceleration due to gravity is 9.8 m/s^2, we can use the formula:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s at the highest point)
u = initial velocity (490 m/s)
a = acceleration (-9.8 m/s^2)
s = displacement (height of highest point)
Rearranging the equation to solve for s:
s = (v^2 - u^2) / (2a)
s = (0 - (490 m/s)^2) / (2 * (-9.8 m/s^2))
s = 12027.55 meters
Therefore, the height of the highest point reached is approximately 12027.55 meters.
B. To find the time required to reach this point, we can use the equation:
v = u + at
Where:
v = final velocity (0 m/s at the highest point)
u = initial velocity (490 m/s)
a = acceleration (-9.8 m/s^2)
t = time
Rearranging the equation to solve for t:
t = (v - u) / a
t = (0 - 490 m/s) / (-9.8 m/s^2)
t ≈ 50 seconds
Therefore, it takes approximately 50 seconds to reach the highest point.
C. The total time before the bullet returns to the ground is twice the time taken to reach the highest point, since the ascent and descent times are equal:
Total time = 2 * t
Total time ≈ 2 * 50 seconds
Total time ≈ 100 seconds
Therefore, the total time before the bullet returns to the ground is approximately 100 seconds.
D. The speed of the bullet upon hitting the ground can be determined using the initial velocity and the total time. The final velocity can be calculated using the equation:
v = u + at
Where:
v = final velocity (unknown)
u = initial velocity (490 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = total time (100 seconds)
Rearranging the equation to solve for v:
v = u + at
v = 490 m/s + (-9.8 m/s^2) * 100 seconds
v = 490 m/s + (-980 m/s^2)
v ≈ -490 m/s
Please note that the negative sign indicates that the speed upon hitting the ground is in the opposite direction of the initial velocity. So, the speed of the bullet upon hitting the ground is approximately 490 m/s, but in the opposite direction. Keep in mind that this calculation is based on the assumption of no air resistance.
To solve this problem, we can use the equations of motion for an object in free fall. Since the bullet is fired directly upward, we can still use these equations because the only force acting on the bullet is gravity.
Before we start, let's define some variables:
- Initial velocity of the bullet, u = 490 m/s (upward)
- Final velocity of the bullet, v = ?
- Acceleration due to gravity, g = 9.8 m/s^2 (downward)
- Height of the highest point reached, h = ?
- Time required to reach this point, t1 = ?
- Total time before the bullet returns to the ground, t2 = ?
- Speed of the bullet upon hitting the ground, v2 = ?
Now let's solve for each part of the problem:
A. Height of the highest point reached (h):
At the highest point, the final velocity of the bullet is 0 m/s since it momentarily comes to rest before falling down. We can use the equation of motion: v^2 = u^2 + 2gh, where v = 0.
0^2 = (490 m/s)^2 + 2 * (-9.8 m/s^2) * h
Solving for h, we get:
9800h = (490 m/s)^2
h = (490 m/s)^2 / 9800 m/s^2
h = 24.5 m
Therefore, the height of the highest point reached is 24.5 meters.
B. Time required to reach this point (t1):
We can use the equation: v = u + gt, where v = 0 m/s (at the highest point).
0 m/s = 490 m/s + (-9.8 m/s^2) * t1
Solving for t1, we get:
490 m/s = 9.8 m/s^2 * t1
t1 = 490 m/s / 9.8 m/s^2
t1 = 50 seconds
Therefore, the time required to reach the highest point is 50 seconds.
C. Total time before the bullet returns to the ground (t2):
Since the time it takes for the bullet to reach the highest point is the same as the time it takes to fall back to the ground, the total time is twice the time required to reach the highest point.
t2 = 2 * t1
t2 = 2 * 50 seconds
t2 = 100 seconds
Therefore, the total time before the bullet returns to the ground is 100 seconds.
D. Speed of the bullet upon hitting the ground (v2):
We know that the initial velocity in the vertical direction is 490 m/s, and the final velocity in the vertical direction is v2 m/s.
Using the equation of motion: v = u + gt, where u = 490 m/s, v = v2 m/s, and t = t2 = 100 seconds, we can solve for v2.
v2 = 490 m/s + (-9.8 m/s^2) * 100 seconds
v2 = 490 m/s - 980 m/s
v2 = -490 m/s
Since the bullet will be moving downward upon hitting the ground, the speed will be negative. However, we usually consider the magnitude of speed, so the speed of the bullet upon hitting the ground is 490 m/s.
Therefore, the speed of the bullet upon hitting the ground is 490 m/s.