Question

An inquisitive physics student and mountain climber climbs a 52.0 m cliff that overhangs a calm pool of water. He throws (not drops) two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.05 m/s downward. Hint: Because the motion is only in the downward direction, choose the positive y-axis to point downward.(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 they are to hit simultaneously?(downward)(c) What is the speed of each stone at the instant the two hit the water?first stone (downward)second stone (downward)

Answer 1

Answer:

Part a)

$t = 3.06 s$

Part b)

$v_2 = 15.22 m/s$

Part c)

$v_{2f} = 35.4 m/s$

Explanation:

Part a)

as we know that speed of the stone is 2.05 m/s

displacement of the stone is 52 m downwards

now we can use kinematics

$d = v_i t + \frac{1}{2}at^2$

$52 = 2.05 t + \frac{1}{2}(9.8)t^2$

$4.9 t^2 + 2.05 t - 52 = 0$

$t = 3.06 s$

Part b)

Since second stone is projected downwards with speed v after time t = 1 s

so relative separation between two stones is given as

$d = vt + \frac{1}{2}at^2$

$d = (2.05)(1.00) + \frac{1}{2}(9.8)(1^2)$

$d = 6.95 m$

so now we can say that if both stone hit the water simultaneously so here second stone will approach 1st atone after t = 3.06 - 1 = 2.06 s

so we have

$v_{rel} t = d_{rel}

[tex](v_2 - v_1) \times (2.06)= 6.95$

here v1 is the speed of first stone after t = 1 s

$v_1 = 2.05 + 9.8(1) = 11.85 m/s$

now we will have

$v_2 - 11.85 = \frac{6.95}{2.06}$

$v_2 = 15.22 m/s$

Part c)

speed of first atone when it hit the water

$v_{1f} = v_1 + at$

$v_{1f} = 2.05 + (9.81)(3.06)$

$v_{1f} = 32.1 m/s$

speed of 2nd stone when it will hit the water

$v_{2f} = v_2 + at$

$v_{2f} = 15.22 + (9.81)(2.06)$

$v_{2f} = 35.4 m/s$