Question

an inquisitive physics student and mountain climber climbs a 53.0 - m - high cliff that overhangs a calm pool of water. he throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. the first stone has an initial speed of 2.12 m/s. (a) how long after release of the first stone do the two stones hit the water? (b) what initial velocity must the second stone have if the two stones are to hit the water simultaneously? magnitude m/s direction (c) what is the speed of each stone at the instant the two stones hit the water? first stone m/s second stone m/s

Explanation:

Step1: Find time for first - stone to reach water

Use the kinematic equation $y = y_0+v_0t+\frac{1}{2}at^2$. Here, $y = 0$ (final position at water level), $y_0=53.0$ m (initial height), $v_0 = 2.12$ m/s (initial velocity) and $a=- 9.8$ m/s² (acceleration due to gravity). So, $0 = 53.0+2.12t-\frac{1}{2}\times9.8t^2$. Rearranging gives $4.9t^2 - 2.12t - 53.0=0$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ where $a = 4.9$, $b=-2.12$ and $c=-53.0$.
$t=\frac{2.12\pm\sqrt{(-2.12)^{2}-4\times4.9\times(-53.0)}}{2\times4.9}=\frac{2.12\pm\sqrt{4.4944 + 1038.8}}{9.8}=\frac{2.12\pm\sqrt{1043.2944}}{9.8}=\frac{2.12\pm32.299}{9.8}$. We take the positive root $t=\frac{2.12 + 32.299}{9.8}\approx3.51$ s.

Step2: Find time for second - stone to reach water

The second - stone is thrown 1.00 s later and reaches the water at the same time as the first. So the time of flight for the second stone, $t_2=t - 1.00=3.51 - 1.00 = 2.51$ s.

Step3: Find initial velocity of second stone

Again using $y = y_0+v_0t+\frac{1}{2}at^2$ with $y = 0$, $y_0 = 53.0$ m, $t = 2.51$ s and $a=-9.8$ m/s².
$0=53.0+v_0\times2.51-\frac{1}{2}\times9.8\times(2.51)^2$.
$0=53.0 + 2.51v_0-30.87$.
$2.51v_0=30.87 - 53.0=-22.13$.
$v_0=\frac{-22.13}{2.51}\approx - 8.82$ m/s. The negative sign indicates the downward direction.

Step4: Find final velocities of both stones

For the first stone, use $v = v_0+at$. Here, $v_0 = 2.12$ m/s, $a=-9.8$ m/s² and $t = 3.51$ s. So $v_1=2.12-9.8\times3.51=2.12 - 34.398=-32.28$ m/s.
For the second stone, $v_0=-8.82$ m/s, $a=-9.8$ m/s² and $t = 2.51$ s. So $v_2=-8.82-9.8\times2.51=-8.82 - 24.598=-33.42$ m/s.

Answer:

(a) $3.51$ s
(b) Magnitude: $8.82$ m/s, Direction: Downward
(c) First stone: $32.28$ m/s, Second stone: $33.42$ m/s