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A player throw a ball upward with initial speed of 30m/s what is time when the ball to come back into his hands?
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A player throw a ball upward with initial speed of 30m/s what is time ...
V=u+at higest point v=030-10Γ—t 30=10tt=3 taken in highest point and total time 3+3=6 second
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A player throw a ball upward with initial speed of 30m/s what is time ...
Calculating the Time for the Ball to Return


To calculate the time it takes for the ball to return to the player's hands, we can use the following kinematic equation:


y = v0t - 0.5gt2



  • y represents the displacement of the ball (in this case, the height it reaches before returning to the player's hands)

  • v0 represents the initial velocity of the ball

  • t represents the time it takes for the ball to reach its maximum height and return to the player's hands

  • g represents the acceleration due to gravity, which is approximately 9.8 m/s2



Plugging in the Values


Using the given initial velocity of 30 m/s, we can plug in the other values and solve for t:


y = v0t - 0.5gt2

0 = (30 m/s)t - 0.5(9.8 m/s2)t2


Simplifying this equation, we get:


4.9t2 - 30t = 0


Factoring out t, we get:


t(4.9t - 30) = 0


This equation has two solutions: t = 0 (which represents the initial time when the ball is thrown) and t = 6.12 seconds (which represents the time when the ball returns to the player's hands).


Conclusion


Therefore, the time it takes for the ball to return to the player's hands is approximately 6.12 seconds. This calculation assumes that air resistance is negligible and that the ball is thrown straight up (i.e. the angle of projection is 90 degrees). In reality, these factors may affect the actual time it takes for the ball to return to the player's hands.
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A player throw a ball upward with initial speed of 30m/s what is time when the ball to come back into his hands?
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