A bullet was fired horizontally from the top of a 1.5x10^2 m tall tower at 2.8x10^2 m/s. Determine its speed on impact with the ground below.
Two masses connected by a rope over a pulley are released, producing the bar graphs below. It took 0.80 seconds for Mass A to reach the top and Mass B to reach the bottom. What was the acceleration of Mass A?
2.05 m/s2
The correct answer is B. 3.20 m/s^2
To determine the speed of the bullet on impact with the ground below, we can use the principle of projectile motion.
In this case, the bullet is fired horizontally. This means that the initial vertical velocity of the bullet is zero, as it is not moving up or down when it is fired. The only initial velocity component to consider is the horizontal component.
We can begin by finding the time it takes for the bullet to reach the ground. We can use the formula for time of flight in vertical motion:
t = √(2h / g)
Where:
t = time of flight
h = vertical height (1.5x10^2 m in this case)
g = acceleration due to gravity (approximately 9.8 m/s^2)
Plugging in the values, we get:
t = √(2 * 150 / 9.8)
t = √(300 / 9.8)
t ≈ √30.61
Since time cannot be negative, the square root of 30.61 will be the positive value of time:
t ≈ 5.53 seconds
Now, we can determine the horizontal distance covered by the bullet using the formula:
d = v * t
Where:
d = horizontal distance
v = horizontal velocity (2.8x10^2 m/s in this case)
t = time of flight (5.53 seconds)
Plugging in the values, we get:
d = (2.8x10^2 m/s) * (5.53 seconds)
d = 1548 m
So the bullet covers a horizontal distance of approximately 1548 meters.
Finally, to find the speed of the bullet on impact, we can use the formula for final velocity:
v_final = √(v_initial^2 + 2ad)
Where:
v_initial = initial velocity (horizontal velocity of the bullet)
a = acceleration (in this case, it is just the acceleration due to gravity, which is 9.8 m/s^2)
d = horizontal distance covered by the bullet (1548 m in this case)
Plugging in the values, we get:
v_final = √((2.8x10^2)^2 + 2 * 9.8 * 1548)
v_final = √(78400 + 30296)
v_final = √108696
v_final ≈ 329.6 m/s
So, the speed of the bullet on impact with the ground below is approximately 329.6 m/s.