Question

An inquisitive physics student and mountain climber climbs a 50m cliff that overhangs a calm pool of water. He throws two stones vertically downwards 1s apart and observes that they cause a single splash. The first stone has an initial velocity of 2ms^-1.
(a) At what time after release of the first stone will the two stones hit the water?
(b) What is the initial velocity that the second stone should have if they are to hit simultaneously?
(c) What will the velocity of each stone be at the instant they hit the water?

Answer 1

Answer on Question #73875, Physics / Mechanics | Relativity

Question. An inquisitive physics student and mountain climber climbs a $S = 50 \, \text{m}$ cliff that overhangs a calm pool of water. He throws two stones vertically downwards $t_0 = 1 \, \text{s}$ apart and observes that they cause a single splash. The first stone has an initial velocity of $v_{01} = 2 \, \text{m/s}$.

(a) At what time $\Delta t$ after release of the first stone will the two stones hit the water?

(b) What is the initial velocity $v_{02}$ that the second stone should have if they are to hit simultaneously?

(c) What will the velocity $v_{1}, v_{2}$ of each stone be at the instant they hit the water?

Solution.

For the first stone

$$S = v_{01} \Delta t + \frac{g \Delta t^{2}}{2}.$$

For the second stone

$$S = v_{02} (\Delta t - t_{0}) + \frac{g (\Delta t - t_{0})^{2}}{2}.$$

(a) At what time $\Delta t$ after release of the first stone will the two stones hit the water?

$$S = v_{01} \Delta t + \frac{g \Delta t^{2}}{2} \rightarrow 50 = 2 \Delta t + 9.8 \cdot \frac{\Delta t^{2}}{2} \rightarrow 9.8 \Delta t^{2} + 4 \Delta t - 100 = 0 \rightarrow \Delta t \approx 3 \, \text{s}.$$

(b) What is the initial velocity $v_{02}$ that the second stone should have if they are to hit simultaneously?

$$S = v_{02} (\Delta t - t_{0}) + \frac{g (\Delta t - t_{0})^{2}}{2} \rightarrow 50 = v_{02} (3 - 1) + \frac{9.8 \cdot (3 - 1)^{2}}{2} \rightarrow v_{02} = 15.2 \, \frac{\text{m}}{\text{s}}.$$

(c) What will the velocity $v_{1}, v_{2}$ of each stone be at the instant they hit the water?

$$v_{1} = v_{01} + g \Delta t = 2 + 9.8 \cdot 3 = 31.4 \, \frac{\text{m}}{\text{s}}.$$ $$v_{2} = 15.2 + 9.8 \cdot (3 - 1) = 34.8 \, \frac{\text{m}}{\text{s}}.$$

Answer. $\Delta t \approx 3 \, \text{s}; v_{02} = 15.2 \, \text{m/s}; v_{1} = 31.4 \, \frac{\text{m}}{\text{s}}; v_{2} = 34.8 \, \frac{\text{m}}{\text{s}}$.

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