Thank you for visiting A rifle is aimed horizontally at a target 45 0 m away The bullet hits the target 1 50 cm below the aim point What. 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 :
To find the bullet's flight time, we can use the kinematic equations that describe the motion of an object under constant acceleration. The horizontal motion does not affect the time it takes to fall vertically since horizontal and vertical motions are independent of each other. So the bullet's flight time is approximately 0.0553 seconds.
The relevant equation for vertical displacement under constant acceleration starting from rest is:
[tex]\[ y = \frac{1}{2} g t^2 \][/tex]
Here, [tex]\( y \)[/tex] is the vertical displacement in meters, [tex]\( g \)[/tex] is the acceleration due to gravity [tex](\( 9.81 \text{ m/s}^2 \))[/tex], and [tex]\( t \)[/tex] is the time in seconds. We need to find [tex]\( t \)[/tex] when [tex]\( y = 1.50 \text{ cm} \)[/tex]. Since the bullet is fired horizontally, we can assume that its vertical motion is solely due to the acceleration due to gravity, which is approximately [tex]\( 9.81 \text{ m/s}^2 \)[/tex].
However, to use the equation, we need to have all the units consistent, so we will first convert 1.50 cm to meters:
[tex]\[ 1.50 \text{ cm} = 0.015 \text{ m} \][/tex]
Now, we can plug our known values into the equation and solve for [tex]\( t \)[/tex]:
[tex]\[ 0.015 \text{ m} = \frac{1}{2} (9.81 \text{ m/s}^2) t^2 \][/tex]
To solve for [tex]\( t^2 \)[/tex], we multiply both sides by 2 and divide by [tex]\( g \)[/tex]:
[tex]\[ t^2 = \frac{2 \times 0.015 \text{ m}}{9.81 \text{ m/s}^2} \][/tex]
[tex]\[ t^2 = \frac{0.03 \text{ m}}{9.81 \text{ m/s}^2} \][/tex]
[tex]\[ t^2 \approx \frac{0.03 \text{ m}}{9.81 \text{ m/s}^2} \approx 0.003058 \text{ s}^2 \][/tex]
To find [tex]\( t \)[/tex], we will take the square root of both sides:
[tex]\[ t \approx \sqrt{0.003058 \text{ s}^2} \][/tex]
[tex]\[ t \approx 0.0553 \text{ s} \][/tex]
So the bullet's flight time is approximately 0.0553 seconds.
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