A rock climber stands on top of a 50-m-high cliff...

A rock climber stands on top of a 50-m-high cliff overhanging a pool of water. He throws two stones vertically downward $1.0 \mathrm{s}$ apart and observes that they cause a single splash. The initial speed of the first stone was $2.0 \mathrm{m} / \mathrm{s}$. a. How long after the release of the first stone does the second stone hit the water? b. What was the initial speed of the second stone? c. What is the speed of each stone as it hits the water?

A rock climber stands on top of a 50-m-high cliff overhanging a pool of water. He throws two stones vertically downward $1.0 \mathrm{s}$ apart and observes that they cause a single splash. The initial speed of the first stone was $2.0 \mathrm{m} / \mathrm{s}$.
a. How long after the release of the first stone does the second stone hit the water?
b. What was the initial speed of the second stone?
c. What is the speed of each stone as it hits the water?

Key Concepts

Relative Timing in Sequential Launches

This concept deals with analyzing motions of objects that are launched or released at different times. By accounting for a time delay between the launches, one can set up equations that allow the synchronization of events (for example, ensuring both stones hit the water simultaneously), which is crucial in solving problems where multiple objects interact with the same event at different starting times.

Solving Quadratic Equations in Kinematics

Many kinematics problems result in quadratic equations when using the equations of motion to solve for time or initial velocity. Being able to recognize, set up, and solve these quadratic equations is a key skill, as it allows for determining the exact moments when certain events occur, such as an object reaching the ground.

Kinematics under Constant Acceleration

This concept involves analyzing the motion of objects which are subject to a constant acceleration, such as gravity. It uses equations like s = ut + ½at² and v = u + at to determine an object's position and velocity at any given time, forming the backbone of solving problems involving free fall or vertical projectile motion.

Equations of Motion

These are the mathematical relationships that describe the movement of an object in terms of its initial velocity, acceleration, time, displacement, and final velocity. They are essential in determining when and how an object reaches a certain point, which in this context is used to calculate the time it takes for each stone to hit the water and their speeds upon impact.

Transcript

0:00 This problem.

00:01 We're gonna talk about gravitational acceleration.

00:04 Remember first, that the gravitational acceleration is 9.8 meters per second squared.

00:11 Also, remember that if we set up a set of a coordinate system such that the y axis points upwards, then the gravitation exploration points downwards towards the center of the earth and the position why off the particle undergoing gravitation acceleration is equal to the initial position.

00:33 Why zero close the initial velocity v zero times of time.

00:37 T minus g over to xi squared, where the minus sign is here because jesus pointing downwards.

00:47 And what we have in our problem is a climber who is on the top off a cliff that has a height of 50 m.

00:59 So i suppose that this here is the cliff.

01:05 You're yeah.

01:06 Okay.

01:07 Let me actually dry it more like this.

01:09 So the primaries on the top here and his throws to rocks dollars towards the water.

01:20 The first brock yeah, uh, and and the first rocket is thrown with an initial speed v zero off 2 m per second.

01:31 And then he throws a second rock with ah different initial speed.

01:40 And the two rocks hit the water at the same time.

01:46 And we know that the second rock was thrown one second after the first rock.

01:56 Okay, so in question a, we want to find how long after the first rock was thrown the second what? the second rock hits the water.

02:08 Notice that the two rocks hit the water at the same time, meaning that the time that we have to calculate here is just the time that it takes for the first rock to hit the water.

02:20 I'm gonna use the equation for the white position.

02:23 I'm gonna choose the origin of my coordinate system to be at the water level, meaning that the initial position of the rock is equal to 50 m in the initial velocity.

02:37 V zero is minus 2 m per second and the minuses here because the climate is throwing the rock downwards.

02:49 Okay, so we have that zero.

02:52 That is the position of the rock.

02:53 When it hits, the water is equal to 50 m.

02:56 The initial position minus two times t minus g.

03:02 That's 9.8.

03:04 Everything here is in the international system of units times t squared.

03:11 So this is the equation that we have to solve 4.90 square plus two t minus 50.

03:23 So let's solve it.

03:25 T is equal to minus two.

03:27 I'm just going to use bhaskar here, plus or minus the square root of two squared minus four times 9.8 times minus 50.

03:41 Uh huh.