a bullet is fired from at a shooting range. The bullet hits the ground after .32 seconds. How far did it travel horizontally and vertically in this time if it was fired at a velocity of 1100 m/s?
To determine the horizontal and vertical distance traveled by the bullet, we can use the equations of motion. In this case, we have the initial velocity of the bullet, the time it is in the air, and we want to find the horizontal and vertical distances.
First, let's consider the vertical motion. We can use the equation of motion for vertical displacement:
𝑦 = 𝑢𝑦𝑡 + 0.5𝑔𝑡²
Where:
𝑦 is the vertical displacement
𝑢𝑦 is the initial vertical velocity (in this case, the bullet is fired horizontally, so 𝑢𝑦 = 0)
𝑔 is the acceleration due to gravity, approximately 9.8 𝑚/𝑠²
𝑡 is the time the bullet is in the air (0.32 s)
Substituting the values into the equation, we get:
𝑦 = 0 + 0.5(9.8)(0.32)²
Simplifying:
𝑦 = 0.5(9.8)(0.1024)
𝑦 = 0.49904
So, vertically, the bullet traveled approximately 0.49904 meters.
Now, let's consider the horizontal motion. The horizontal velocity of the bullet remains constant throughout its flight because there is no horizontal force acting on it. So, we can use the equation:
𝑥 = 𝑢𝑥𝑡
Where:
𝑥 is the horizontal displacement
𝑢𝑥 is the initial horizontal velocity (in this case, 1100 𝑚/𝑠)
𝑡 is the time the bullet is in the air (0.32 s)
Substituting the values into the equation, we get:
𝑥 = 1100(0.32)
𝑥 = 352
So, horizontally, the bullet traveled approximately 352 meters.
To summarize, the bullet traveled approximately 0.49904 meters vertically and 352 meters horizontally in 0.32 seconds.