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A ball is thrown downward with an initial velocity of 15 m/s. Using the approximate value of g = 10 m/s². Calculate the velocity of the ball 3.2 seconds after it is released.



Answer :

Answer:47m/sec

Explanation:

To solve this problem, we can use the kinematic equation that relates the final velocity (vv), initial velocity (uu), acceleration due to gravity (gg), and time (tt):

v=u+gt

Given:

   Initial velocity, u=15u=15 m/s (downward)

   Acceleration due to gravity, g=10g=10 m/s22 (downward)

   Time, t=3.2t=3.2 seconds

Now, substitute these values into the equation:

v=15 m/s+(10 m/s2)(3.2 s)v=15m/s+(10m/s2)(3.2s)

Calculate the right-hand side:

v=15 m/s+32 m/sv=15m/s+32m/s

v=47 m/sv=47m/s

Therefore, the velocity of the ball 3.2 seconds after it is released is 47 m/s47m/s​ downward.

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