Physics: Maximum Height Calculation

This problem involves calculating the maximum height reached by an object thrown vertically upwards. We use the principles of kinematics.

Applying Kinematic Equations

• Identify Given Values:
  • Initial velocity, $ u = 10\ m/s $.
  • Acceleration due to gravity, $ g = 9.8\ m/s^2 $. Since the ball is moving upwards, gravity acts downwards, so the acceleration $ a = -g = -9.8\ m/s^2 $.
  • Final velocity at maximum height, $ v = 0\ m/s $. This is because the ball momentarily stops before falling back down.
• Identify Unknown Value:
  • Maximum height, $ H $ (which is the displacement $ s $).
• Select Relevant Formula: The kinematic equation that relates initial velocity ($ u $), final velocity ($ v $), acceleration ($ a $), and displacement ($ s $) is:

  $ v^2 = u^2 + 2as $

• Adapt Formula for Maximum Height: Substitute $ v = 0 $, $ a = -g $, and $ s = H $:

  $ 0^2 = u^2 + 2(-g)H $

  $ 0 = u^2 - 2gH $

• Solve for Maximum Height (H): Rearrange the equation:

  $ 2gH = u^2 $

  $ H = \frac{u^2}{2g} $

• Substitute Values and Calculate:

  $ H = \frac{(10\ m/s)^2}{2 \times (9.8\ m/s^2)} $

  $ H = \frac{100\ m^2/s^2}{19.6\ m/s^2} $

  $ H \approx 5.102\ m $

• Final Answer: Rounding to two decimal places, the maximum height reached is approximately $ 5.10\ m $.