Thank you for visiting A person skateboarding with a constant speed of 1 70 m s releases a ball from a height of 1 65 m above the ground. 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 :
The speed of the ball at time t = 0.200 s can be found using the equation for motion under constant acceleration. The ball is released from a height of 1.65 m above the ground, which means that its initial velocity is 0 m/s.
The acceleration due to gravity is approximately 9.8 m/s^2. Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can substitute the known values into the equation.
The initial velocity u is 0 m/s, the acceleration a is -9.8 m/s^2 (negative because it acts in the opposite direction to the initial velocity), and the time t is 0.200 s. The equation becomes v = 0 + (-9.8) * 0.200.
The equation for motion under constant acceleration is v = u + at. In this case, the ball is released from rest (u = 0) and experiences an acceleration due to gravity of -9.8 m/s^2. The negative sign is used because the acceleration acts in the opposite direction to the initial velocity. To find the speed of the ball at time t = 0.200 s, we substitute the known values into the equation. The acceleration is -9.8 m/s^2 and the time is 0.200 s. Substituting these values, we get v = 0 + (-9.8) * 0.200. Simplifying the equation, we have v = -1.96 m/s. Therefore, the speed of the ball at time t = 0.200 s is -1.96 m/s. The negative sign indicates that the ball is moving downward.
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