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A student drops a ball from a height of  . The acceleration of the ball due to gravity is  .How long will it take the ball to reach the ground?

Question

A student drops a ball from a height of  . The acceleration of the ball due to gravity is  .How long will it take the ball to reach the ground?

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Solution

Sure, let's solve this step by step.

  1. Identify the given values:

    • Initial height (h) = h h
    • Acceleration due to gravity (g) = g g
    • Initial velocity (u) = 0 (since the ball is dropped)
  2. Use the kinematic equation for motion under constant acceleration: h=ut+12gt2 h = ut + \frac{1}{2}gt^2

  3. Substitute the known values into the equation: h=0t+12gt2 h = 0 \cdot t + \frac{1}{2}gt^2 Simplifying, we get: h=12gt2 h = \frac{1}{2}gt^2

  4. Solve for time (t): t2=2hg t^2 = \frac{2h}{g} t=2hg t = \sqrt{\frac{2h}{g}}

So, the time it will take for the ball to reach the ground is: t=2hg t = \sqrt{\frac{2h}{g}}

This problem has been solved

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