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A ball is thrown horizontally from a height of 20m. What is the time it takes to hit the ground?

Physics

ScanSolve AI Answer

Answer

2 s\text{2 s}

Step-by-step solution

3 steps

To determine the time taken by the ball to strike the surface of piston A, we only need to consider the vertical motion of the ball.

Given:
Initial vertical velocity, uy=0 m/su_y = 0 \text{ m/s} (since the ball is thrown horizontally).
Vertical distance (height), h=20 mh = 20 \text{ m}.
Acceleration due to gravity, g=10 m/s2g = 10 \text{ m/s}^2 (standard approximation).

We use the kinematic equation for vertical displacement:

sy=uyt+12gt2s_y = u_y t + \frac{1}{2} g t^2
  1. Substitute the given values into the equation.

    20 m=(0 m/s)t+12(10 m/s2)t220 \text{ m} = (0 \text{ m/s}) t + \frac{1}{2} (10 \text{ m/s}^2) t^2 20=0+5t220 = 0 + 5 t^2 20=5t220 = 5 t^2
  2. Solve for t2t^2.

    t2=205t^2 = \frac{20}{5} t2=4t^2 = 4
  3. Solve for tt.

    t=4t = \sqrt{4} t=2 st = 2 \text{ s}

    The time taken by the ball to strike the surface of piston A is 2 s\boxed{\text{2 s}}.

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