Thank you for visiting A rifle is aimed horizontally at a target 54 m away The bullet hits the target 2 4 cm below the aim point What was. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Final answer:
The bullet's flight time was roughly 0.07 seconds and the speed it left the barrel was approximately 771.43 meters per second.
Explanation:
To solve this problem, we can use the principles of physics related with projectile motion. This means we treat the horizontal and vertical components of the bullet's motion separately. We can ignore the horizontal motion when calculating the time of flight, as it does not affect the vertical motion.
Knowing that the bullet falls 2.4 cm (which is 0.024 m when converted meter) below the aim point, we're dealing with a controlled form of free fall in which the only force acting on the bullet after it leaves the rifle is gravity with acceleration due to gravity g = 9.8 m/s².
From the kinematic equation of motion: h = 0.5gt², where h is the height fallen by the bullet (2.4 cm = 0.024 m), and t is the time of flight we want to find. By solving this equation for t, we find:
t = sqrt((2xh)/g) = sqrt((2x0.024 m)/(9.8 m/s²)) = 0.07 s
The horizontal distance travelled by the bullet (54 m) during this time is known as the range (R) of the projectile which can be calculated using R = vt, where v is the bullet's speed and t is the flight time we just calculated. Solving for the bullet speed v:
v = R/t = 54 m / 0.07 s = 771.43 m/s
Learn more about Projectile Motion here:
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