From three points on a vertical shore, three identical balls…
From three points on a vertical shore, three identical balls are launched horizontally at the same moment with speeds $v_{01} = 50$, $v_{02} = 75$ and $v_{03} = 100\ \mathrm{m/s}$. The first ball hits the water at a horizontal distance $D_1 = 100\ \mathrm{m}$. All three hit the water at the same time. Find (a) the flight times; (b) the launch heights; (c) the speeds on impact.
Explanation of Method
Three balls launched horizontally that land together.
Step 1: (a) Flight time
Horizontally each ball moves uniformly. For the first: \[t_1 = \frac{D_1}{v_{01}} = \frac{100}{50} = 2\ \mathrm{s}.\]
All three land at the same moment and were launched at the same moment, so
\[t_1 = t_2 = t_3 = 2\ \mathrm{s}\]
Step 2: (b) Heights
The fall time depends only on the height (the horizontal speed does not matter), so equal times mean equal heights: $h = \tfrac12 g t^2 = \tfrac12 \cdot 9.81 \cdot 4$.
\[h_1 = h_2 = h_3 = 19.62\ \mathrm{m}\]
Step 3: (c) Speeds on impact
All get the same vertical speed $gt = 19.62\ \mathrm{m/s}$: $v_i = \sqrt{v_{0i}^2 + 19.62^2}$.
\[v_1 \approx 53.7\ \mathrm{m/s}, \quad v_2 \approx 77.5\ \mathrm{m/s}, \quad v_3 \approx 101.9\ \mathrm{m/s}\]
- 50 m/s
- 75 m/s
- 100 m/s
Result
\[\boxed{\,t_1 = t_2 = t_3 = 2\ \mathrm{s}\,}\]
\[\boxed{\,h_1 = h_2 = h_3 = 19.62\ \mathrm{m}\,}\]
\[\boxed{\,v_1 \approx 53.7\ \mathrm{m/s}, \quad v_2 \approx 77.5\ \mathrm{m/s}, \quad v_3 \approx 101.9\ \mathrm{m/s}\,}\]
Remark
About the wording. Three points on a vertical shore suggests different heights, which would contradict "same time". A common variant of this problem says the balls land at the same place (100 m out); then the times are 2, 1.33 and 1 s, the heights 19.62, 8.72 and 4.91 m, and the impact speeds 53.7, 76.1 and 100.5 m/s. (The original value $v_1 = 53.77$ is a small rounding slip.)