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Exactly! The 3rd EOM says that \(v^2 = u^2 + 2*a*s\). We know that \(u = 10 \,\mathrm{m/s}\) and \(s = 5 \,\mathrm{m}\). When it reaches the maximum height, the ball comes to a stop. So, \(v = 0 \,\mathrm{m/s}\).
Imagine that you drop a ball from a building. In Physics, when we say drop something, we mean that the initial velocity is \(0\). If the ball takes \(5 \, \mathrm{s}\) to reach the ground and the ball is accelerating downwards at \(10 \, \mathrm{m/s^2}\), which of the following would you use to determine the height of the building?

1st EOM: \(v = u + a*t\). 2nd EOM: \(s = u*t + \frac{1}{2}*a*t^2\)
  • 1st EOM with \(u = 0 \, \mathrm{m/s}\), \(a = -10 \, \mathrm{m/s^2}\) and \(t = 5 \, \mathrm{s}\)
  • 2nd EOM with \(u = 0 \, \mathrm{m/s}\), \(t = 5 \, \mathrm{s}\) and \(a = -10 \, \mathrm{m/s^2}\)
  • 3rd EOM with \(u = 0 \, \mathrm{m/s}\), \(t = 5 \, \mathrm{s}\) and \(a = -10 \, \mathrm{m/s^2}\)
  • 2nd EOM with \(u = 10 \, \mathrm{m/s}\) and \(t = 5 \, \mathrm{s}\)