From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. The bullet puts a hole in a window of another building and hits the wall that faces the window. (y = 0.42 m, and x = 7.2 m.)Determine the distances D and H, which locate the point where the gun was fired. Assume that the bullet does not slow down as it passes through the window.
let vy=is vertical componet then t' time of flight
x=ut'
y=vyt'-gt'^2/2
then eliminate the two equation
H= y +gt^2/2
and to find vy
vy=yu/x +gx/2u
D+x=ut
substitute the tow equation,
D=usqrt(2H)/g -x
To determine the distances D and H, which locate the point where the gun was fired, we can use the kinematic equations of motion.
Let's break down the given information:
- The bullet leaves the gun at a speed of 340 m/s parallel to the ground.
- The bullet hits the wall that faces the window, and we are given the coordinates of the impact point: y = 0.42 m and x = 7.2 m.
We can consider two separate motions: the horizontal motion (along the x-axis) and the vertical motion (along the y-axis).
1. Vertical Motion:
In this case, the bullet undergoes free-fall motion vertically, as it is not influenced by any horizontal forces. We can use the equation:
y = (1/2)gt^2
where y is the vertical displacement, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken.
Since the bullet is fired parallel to the ground, the time taken to hit the wall is the same as the time taken to fall 0.42 m vertically. Hence, we can rearrange the equation to solve for time:
0.42 = (1/2) * 9.8 * t^2
Solving for t:
0.42 = 4.9t^2
t^2 = 0.42/4.9
t^2 ≈ 0.0857
t ≈ √0.0857
t ≈ 0.293 s
2. Horizontal Motion:
In this case, the bullet moves with a constant velocity of 340 m/s along the x-axis. The horizontal distance traveled (x) is given as 7.2 m. We can use the equation:
x = vt
where x is the horizontal displacement, v is the horizontal velocity, and t is the time taken.
Substituting the given values:
7.2 = 340 * t
t = 7.2 / 340
t ≈ 0.0212 s
Now we have the time taken for both vertical and horizontal motions.
3. Finding Distances D and H:
Using the known time values, we can calculate the distances D and H.
D = v * t
D = 340 * 0.0212
D ≈ 7.208 m
H = g * t^2 / 2
H = 9.8 * (0.293^2) / 2
H ≈ 0.423 m
Therefore, the distances D and H are approximately 7.208 m and 0.423 m, respectively. These distances locate the point where the gun was fired.