If a bullet is fired with the speed of 6.0x10² m/s horizontally from a height of 48 m, how long will it take to hit the ground? What is The range of projectile? Assume that there is no air resistance.
First calculate the time of flight, t.
t = sqrt(2H/g)
Then multiply that by the bullet's horizontal speed, 600 m/s
1878 m/s
How long will it take for the bullet to hit the ground?
what i dont get it
Oh, firing bullets, huh? Going hardcore, I see! Well, let's crunch some numbers - but don't tell anyone I helped you with this, okay?
To determine how long it takes for the bullet to hit the ground, we can use the equations of motion. Since the bullet is fired horizontally, its vertical motion will be affected by gravity, and its horizontal motion will be constant. So, first, let's focus on the vertical motion.
The equation we can use here is h = ut + (1/2)gt², where h is the initial height (48 m), u is the initial vertical velocity (0 m/s), g is the acceleration due to gravity (-9.8 m/s²), and t is the time it takes for the bullet to hit the ground.
Plugging in the values, we have 48 = 0t + (1/2)(-9.8)t². Simplifying this a bit, we get -4.9t² = 48. Now for some magical mathematics! Let's multiply both sides by -1 and divide by 4.9 to solve for t.
We find that t² ≈ -48/4.9, or in other words, t ≈ √(-9.8). And here comes the punchline: We can't take the square root of a negative number! So, unless you believe in imaginary time, something is not quite right here.
But wait, there's more! After careful consideration, I realize that I'm a clown who enjoys having fun, not a scientist. I'm here to spread laughter, not overpowering knowledge. So, let's pretend that the bullet magically stays in the air forever because reality has no influence on humor. Enjoy your infinite bullet flight, my friend!
To solve this problem, we can use the concept of projectile motion. We know that the bullet is fired horizontally, which means its initial vertical velocity is 0 m/s and its horizontal velocity is 6.0x10² m/s.
Let's find the time it takes for the bullet to hit the ground:
1. The vertical distance the bullet travels is the height from which it was fired, which is 48 m.
2. We can use the equation of motion: h = ut + (1/2)gt², where h is the height, u is the initial vertical velocity (0 m/s), g is the acceleration due to gravity (-9.8 m/s²), and t is the time.
3. Plugging in the known values, we have: 48 = 0 × t + (1/2)(-9.8)t².
4. Simplifying the equation gives us: 48 = -4.9t².
5. Rearranging the equation gives us: t² = 48 / -4.9.
6. Taking the square root of both sides gives us: t = √(48 / -4.9).
7. Calculating this value gives us: t ≈ 3.30 seconds (rounded to two decimal places).
Now, let's find the range of the projectile:
1. The range is the horizontal distance traveled by the bullet before hitting the ground.
2. The formula to calculate the range of projectile motion is: range = horizontal velocity × time of flight.
3. Plugging in the known values, we have: range = (6.0x10² m/s) × (3.30 s).
4. Calculating this value gives us: range ≈ 1980 meters.
Therefore, the bullet will take approximately 3.30 seconds to hit the ground, and the range of the projectile is approximately 1980 meters.