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 $$t = 5 \, \mathrm{s}$$