Thank you for visiting A projectile is thrown from the top of a 46 m tall building with an initial velocity of 11 m s in the x direction. 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 explain how we arrived at this answer, let's consider the motion of the projectile. The initial velocity of the projectile is given as 11 m/s in the +x direction. We know that the projectile is thrown from the top of a 46 m tall building.
To find the final velocity just before the projectile strikes the ground, we can use the kinematic equation: v^2 = u^2 + 2as- v is the final velocity (what we're trying to find)- u is the initial velocity (11 m/s) - a is the acceleration (in this case, due to gravity, approximately -9.8 m/s^2) - s is the displacement (46 m, since the projectile falls from the top of the building to the ground)
However, it's important to note that the question asks for the speed, not the velocity. Speed is the magnitude of velocity, so it is always positive. Therefore, the final speed of the projectile just before it strikes the ground is approximately 44.7 m/s.
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