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Sagot :
We are given the following information
Initial velocity of ball = 26 m/s
Distance traveled by ball = 9 cm = 9/100 = 0.09 m
We are asked to find the deceleration of the ball.
Recall the third equation of the motion given by
[tex]v^2_f=v^2_i+2as[/tex]Where a is the deceleration, s is the distance traveled, vi is the initial velocity and vf is the final velocity of the ball.
vf = 0 since the ball is stopped after getting caught by the catcher.
Let us substitute the given values into the above equation
[tex]\begin{gathered} v^2_f=v^2_i+2as \\ 0=26^2_{}+2\cdot a\cdot0.09 \\ 0=676+0.18\cdot a \\ 0.18\cdot a=-676 \\ a=-\frac{676}{0.18} \\ a=-3755.56\; \frac{m}{s^2} \end{gathered}[/tex]Therefore, the deceleration of the ball is -3755.56 m/s^2
The negative sign indicates deceleration.
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