A ball is thrown straight up from the edge of the roof of a...

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof $1.00 \mathrm{~s}$ later. Ignore air resistance. (a) If the height of the building is $20.0 \mathrm{~m},$ what must the initial speed of the first ball be if both are to hit the ground at the same time? On the same graph, sketch the positions of both balls as a function of time, measured from when the first ball is thrown. Consider the same situation, but now let the initial speed $v_{0}$ of the first ball be given and treat the height $h$ of the building as an unknown. (b) What must the height of the building be for both balls to reach the ground at the same time if (i) $v_{0}$ is $6.0 \mathrm{~m} / \mathrm{s}$ and (ii) $v_{0}$ is $9.5 \mathrm{~m} / \mathrm{s} ?$ (c) If $v_{0}$ is greater than some value $v_{\max },$ no value of $h$ exists that allows both balls to hit the ground at the same time. Solve for $v_{\max }$. The value $v_{\max }$ has a simple physical interpretation. What is it? (d) If $u_{0}$ is less than some value $v_{\min }$, no value of $h$ exists that allows both balls to hit the ground at the same time. Solve for $v_{\min }$. The value $v_{\min }$ also has a simple physical interpretation. What is it?

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof $1.00 \mathrm{~s}$ later. Ignore air resistance. (a) If the height of the building is $20.0 \mathrm{~m},$ what must the initial speed of the first ball be if both are to hit the ground at the same time? On the same graph, sketch the positions of both balls as a function of time, measured from when the first ball is thrown. Consider the same situation, but now let the initial speed $v_{0}$ of the first ball be given and treat the height $h$ of the building as an unknown.
(b) What must the height of the building be for both balls to reach the ground at the same time if (i) $v_{0}$ is $6.0 \mathrm{~m} / \mathrm{s}$ and (ii) $v_{0}$ is $9.5 \mathrm{~m} / \mathrm{s} ?$ (c) If $v_{0}$ is greater than some value $v_{\max },$ no value of $h$ exists that allows both balls to hit the ground at the same time. Solve for $v_{\max }$. The value $v_{\max }$ has a simple physical interpretation. What is it? (d) If $u_{0}$ is less than some value $v_{\min }$, no value of $h$ exists that allows both balls to hit the ground at the same time. Solve for $v_{\min }$. The value $v_{\min }$ also has a simple physical interpretation. What is it?

Key Concepts

Kinematic Equations under Constant Acceleration

This concept involves using the equations of motion—such as displacement, velocity, and time relations—when acceleration is constant (in these problems, due to gravity). It is crucial for solving problems where position and velocity at different times are determined, particularly in free-fall and projectile motion scenarios.

Projectile Motion in One Dimension

This encompasses the study of objects moving under the influence of gravity, where the motion can be analyzed in vertical (or horizontal) components separately. The technique is used to determine the time evolution of objects thrown upward or dropped, taking into account initial velocities and displacement from the start.

Free Fall and Gravitational Acceleration

This refers to the motion of objects under the sole influence of gravity, with the acceleration 'g' being constant. It is fundamental in determining how long an object takes to fall from a given height and is used to model both the ball dropped (with zero initial velocity) and the ball thrown upward.

Time Delay and Relative Motion

This concept deals with analyzing situations where two objects begin their motion at different times. It is important for setting up conditions (such as both hitting the ground simultaneously) by accounting for the delayed start of one object relative to the other, thereby requiring careful selection of time references in the equations.

Graphical Analysis of Motion

Graphing the positions of objects as a function of time provides a visual representation of their motion. In these types of problems, plotting position versus time helps in understanding how different initial conditions and time delays affect the trajectories and the meeting points (e.g., both reaching the ground simultaneously).

Transcript

00:03 This problem is we have mobility with a high edge, with high edge.

00:13 And we have two balls.

00:15 The first ball will be thrown straight up and get to maximum position and get down.

00:27 This is the first one.

00:29 We have initial velocity here.

00:34 And the second one is after one later from the second later from the first one.

00:40 First one and then we'll free drop from this building and go down to the ground.

00:50 This is the second one.

00:54 This is the first one.

00:57 Then the problem of part a is to want us to find out if the two balls hit the ground at the same time and the height is 20 meters and the what is the velocity of the first one, of the initial velocity of the first ball.

01:24 Okay, this is the problem of part a.

01:32 From this solution, we should understand the equation, the displacement of y will be equal to v0 t minus half g t square.

01:50 Okay, okay, if we set, if we set the ball two need a time, to hit the ground and then the ball one okay would be need the t plus one because the ball two is one second letter from the ball one.

02:32 Then as we know, there are displacements for these two cases, their displacements are the same.

02:42 So that means the delta 4 -1 is equal to delta 2, delta -by -2.

02:53 Therefore, we can say that the 4 -1, there is the initial velocity for the 4 -1 -v -0 and the t plus 1 minus 5g, t plus 1 -2, is equal to 4 -2, the displacement of ball 2.

03:15 Then this will be half g t squared.

03:25 So from here we can get the v0 could be the denominator to the t plus 1, and the numerator could be half g t plus 1 square minus half g t squared so after the arrangement, this equation could be the 5g t plus 1 and 2t plus 1.

04:13 We said the equation to be 1.

04:17 Now, this is the equation to get the initial velocity, but we must know what is t.

04:31 T is the time.

04:34 Okay, is t is time for bought 2.

04:38 Now, how can we find the t? first of all, we understand that for the b2, the vf square could be the 2gh.

04:52 Okay.

04:53 So we know the vf could be the square root 2gh.

05:00 Then we also know that we also know this vf could be the because v0, okay, could be v0, in this condition is 0, so it would be gt.

05:18 Then we know the t could be square root, g2h.

05:30 Now we know the edge for this condition is 20 meters.

05:37 So then we can plug in the number 2 plus 20 and this could be 2 .20 seconds for the t then plug in then substitute this one to equation 1 then we can get the v0 to be 5 9 .8 and the 2 .02 plus 1 2 times 2 .2 then we can get the initial velocity for 1.

06:32 This is a .2 meter per second.

06:41 This is the answer for part a.

06:48 And we can also draw a picture as this one.

06:57 This is position y, and this is t.

07:01 And then the initial the ball one if this is 20 and we can go up and go down and this is the maximum position and this is my maximum position could be 0 .13 second oh no 0 .8 3 second and the 1 second could be here okay and what's the ball 2 if the ball 2 for the first second, it still stays at the 20 meters.

07:47 Then after that, then we will go down to here.

08:00 Okay, this is the graph for the two balls.

08:08 So there is a hint from here.

08:12 If we want to get the two ball to hit the ground at the same time, the first ball at the time equal to 1, the ball must pass over the maximum position of y.

08:33 Then this has the possibility to get the same type to hit the ground for the two balls.

08:43 We will discuss this later.

08:48 Okay, now let's find out the part b.

08:56 What's the part b? the part b is if the we know the initial kind of for the port -a, port 1, and what's the height of the building to get the two ball hit the ground at the same time? okay, for the first case, v0 equal to 0 .206 m per second, and what's the height for this one? okay, now let's calculate this.

09:51 From this equation, a, we can rearrange this equation.

09:57 Could be from one, we can get this equation, v0t plus v0.

10:12 This denominator multiplied for the vlin, v0, and then we can get this one, 2g, 2t, plus 1.

10:30 Then we can get this.

10:38 Then from this one, we can get the t equal to from here, we can get the t could be v0 minus g is equal to half g minus v0.

11:12 And finally, it could be t equal to v0 minus g, half g minus v0.

11:35 This is our equation 3, and this could be our equation 2.

11:56 Okay, here.

12:01 Now, let's find out the t.

12:08 Okay.

12:09 For this one, the case one, the t, plug in the long numbers, v0 right now is 6 .0, 6 minus 0 .9 .8...