A pistol is fired horizontally toward a target 120 m away but at the same height. The bullet's velocity is 200 m/s. How long does it take the bullet to get to the target? How far below the target does the bullet hit?
please explain why you used the formula and how you got the answer.
time=distance/velocity=120/200 seconds
height fell=1/2 g t^2, and your about 5 feet is correct.
thanks
Well, before I answer your question, let me just say that I'm glad you're not using the pistol to shoot the target. That would be a pretty dangerous way of trying to hit something 120 m away!
Now, let's get to the math part. To find out how long it takes for the bullet to reach the target, we can use the formula for time:
Time = Distance / Velocity
In this case, the distance is 120 m and the velocity is 200 m/s. Plugging those values into the formula gives us:
Time = 120 m / 200 m/s = 0.6 seconds
So, it takes the bullet 0.6 seconds to reach the target. That's actually pretty quick!
Now, to find out how far below the target the bullet hits, we need to take gravity into account. Gravity will cause the bullet to drop as it travels horizontally. We can use the formula we learned in physics class:
Distance = 0.5 * acceleration due to gravity * time^2
The acceleration due to gravity is approximately 9.8 m/s^2. Plugging in the value of time we found earlier (0.6 seconds), we get:
Distance = 0.5 * 9.8 m/s^2 * (0.6 s)^2 = 1.764 meters
So, the bullet hits the target 1.764 meters below where it was aimed. That's a pretty significant drop!
I hope that clears things up for you. Remember, it's always better to use humor than a pistol when it comes to hitting targets!
To determine the time it takes for the bullet to reach the target, we can use the formula for displacement:
\[s = ut + \frac{1}{2}at^2\]
where:
- \(s\) is the displacement (in this case, the distance to the target)
- \(u\) is the initial velocity
- \(t\) is the time taken
- \(a\) is the acceleration
Since the bullet is fired horizontally, there is no vertical acceleration. Therefore, \(a = 0\).
To find the time, we can rearrange the formula and solve for \(t\):
\[t = \frac{s}{u}\]
Plugging in the values given, with \(s = 120 \, \text{m}\) and \(u = 200 \, \text{m/s}\), we can calculate:
\[t = \frac{120 \, \text{m}}{200 \, \text{m/s}} = 0.6 \, \text{s}\]
So it takes the bullet 0.6 seconds to reach the target.
Next, to determine how far below the target the bullet hits, we can use the equation for vertical displacement:
\[s_y = ut_y + \frac{1}{2}at_y^2\]
Since the bullet is traveling horizontally, its initial vertical velocity (\(u_y\)) is zero. The formula becomes:
\[s_y = \frac{1}{2}at_y^2\]
We know that the acceleration due to gravity (\(a_y\)) is approximately \(9.8 \, \text{m/s}^2\) downward. We need to find \(t_y\), the time taken in the vertical direction, which is the same as the time taken horizontally (0.6 seconds).
Plugging in the values, we can calculate the vertical displacement (\(s_y\)):
\[s_y = \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (0.6 \, \text{s})^2 = 1.764 \, \text{m}\]
Therefore, the bullet would hit the target 1.764 meters below its original height.
would it be ~
Delta(y)= 1/2(9.8m/s^2)(.6s)^2 = 1.764 so it would be about 5 feet off? That's quite a miss..
Could you specify the 120/200? is it just Delta(t)= 120/200 ? i want to set it up right.
it takes 120/200 = .6 seconds to reach the target.
So, how far does the bullet fall in .6 s?