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Sagot :
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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