A bullet is fired from a rifle 3m above the ground level at a target 200m away. The initial velocity of the bullet is 300 m/s. How low must the center of the target be off the ground for the bullet to successfully hit it?
t=s/v = 200/300 = 0.67 s.
Δy=gt²/2=9.8•0.67²/2 =2.18 m
h=H-Δy=3-2.18= 0.82 m
To find the required height of the target for the bullet to hit it, we can break down the problem into two components: the horizontal and vertical motion of the bullet.
First, let's consider the horizontal motion. The bullet travels 200m horizontally and we know its initial velocity is 300 m/s. We can use the equation:
distance = velocity × time
where the distance is 200m and the velocity is 300 m/s. Solving for time:
time = distance / velocity = 200m / 300 m/s = 2/3 s
Now, let's move on to the vertical motion. The bullet was fired from a height of 3m above the ground level. We want to determine the required height of the target for the bullet to hit it. To calculate this, we need to find how far the bullet drops vertically during the time it takes to travel horizontally.
The vertical motion can be described using the equation:
distance = initial velocity × time + (1/2) × acceleration × time²
where the initial velocity is 0 m/s because the bullet starts and ends at the same vertical position, acceleration is the acceleration due to gravity (-9.8 m/s²), and time is 2/3 s.
Plugging in the values:
3m = 0 × (2/3) s + (1/2) × (-9.8 m/s²) × (2/3 s)²
Simplifying:
3m = (-4.9 m/s²) × (4/9 s²)
Next, solve for the vertical distance (how much the bullet drops):
∆y = (-4.9 m/s²) × (4/9) s²
∆y = -10.89 m
Since the bullet needs to hit the target, the target must be below the initial height of the bullet. Therefore, the center of the target must be at least 10.89 m below the initial height (3 m) for the bullet to successfully hit it.