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 s later. Ignore air resistance. (a) If the height of the building is 20.0 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 m/s and (ii) $v_0$ is 9.5 m/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 $v_0$ 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

Boundary Conditions in Kinematics

Boundary conditions in kinematics refer to the limits within which certain physical scenarios are possible. In problems where objects must meet a condition such as hitting the ground simultaneously, there may be limits on parameters like initial speed or drop height. Maximum or minimum allowable values (such as v_max or v_min) arise from these constraints and have clear physical interpretations, often corresponding to the exact balance between the extra distance climbed or descended and the delay in release.

Graphical Representation of Motion

Graphical analysis in kinematics involves plotting the positions (or velocities) of moving objects as a function of time. This visual representation helps in understanding the trajectories of objects, identifying points of intersection, and verifying conditions such as simultaneous impacts. It provides an intuitive grasp of how changes in initial conditions or release times affect the motion.

Projectile Motion

Projectile motion involves the movement of objects under the influence of gravity alone, assuming negligible air resistance. In vertical motion, the object’s acceleration is constant (typically denoted as g). This subject is fundamental in understanding the way objects travel when they are projected upward or dropped from a height, and it serves as the basis for analyzing the motion parameters such as displacement, velocity, and time of flight.

Kinematic Equations

Kinematic equations relate displacement, velocity, acceleration, and time for objects moving with constant acceleration. These equations are used to predict an object’s future position and velocity based on its initial conditions. In free-fall problems, they help determine quantities like the time it takes for an object to hit the ground, the height reached by an object thrown upward, and the impact velocity.

Simultaneous Impact Condition

The simultaneous impact condition is the requirement that two objects, though released at different times and possibly with different initial velocities, must reach the ground at the same moment. This involves setting the time-of-flight equations of both objects equal to each other and solving for the unknowns. This concept is crucial when comparing trajectories in problems involving delayed launches or different initial conditions.

Transcript

00:01 And part a of this problem, we want to find the initial velocity that we have to give the first ball so that the second ball, which is dropped one second after the first ball is thrown, lands at the exact same time as the first ball.

00:14 Let us denote by y1, the height of the thrown ball, and it is equal to h plus v .0 t minus one -half times gt squared.

00:30 And we'll denote by two the ball that is dropped, and it's equal to 8.

00:34 Minus one half g t minus t not since it's dropped at some t not time later squared and t not for reference is one second now we want them to land at the same time and at the same y value so we're going to set y1 equal to y2 and we're set it to zero we're also going to want to expand this binomial here t minus t not squared and when we we do that we get t squared minus two t not times t plus t not squared and so whenever i equate these two expressions i get h plus v not t minus one half gt squared is equal to h minus one half g and i'm going to substitute in my expanded binomial here so t squared minus two t not times t plus t not squared now i can cancel the h's, and my ultimate goal here is to solve for t.

02:01 And so whenever i distribute this one -half -g across all terms, remember t -not is treated as a constant.

02:08 Really, we just want to solve for t.

02:10 So we're going to move this term with the t over to the other side.

02:14 We're going to cancel this minus one -half g -t -squared with this minus -1 -half g -t -squared.

02:20 And what we get in the end, and i'm skipping some steps here, because otherwise the solution would take too long.

02:28 But after just some algebra, and my pin's messing up here, see if i can fix it, we get t is equal to negative g t0 squared over two times one over v0 minus g t, t0.

02:49 And so given a v0 and given a t not, we can determine the time from that.

02:56 And now we want to substitute this time into the original y1 equation, which is this, plus v -0 -t, minus 1 -half g -t squared.

03:09 We want to substitute this time value in here.

03:13 And then we want to set y1 to be zero, since that's where they're going to meet up.

03:19 And so we get 0 is equal to h plus v -0 times t -0 over 2 times this fraction, 1 over 1 minus v0 over g t not all i did here is i made another simplifying step by dividing both sides by g t not so that removes some stuff here and then also turn this into a 1 and then this into a v0 over g t not and then i have the minus 1 half g t not squared over 4 times 1 over 1 minus v0 over g t not and this is squared since we have t squared there now at this point you would want to get a common denominator between these two terms this one has the bottom squared this one doesn't so you're going to multiply this term by a copy of one over one minus v not over gt not and then once you have a common denominator across these two you're going to combine them and you're going to get some simplification and we ultimately want to solve for the height of this thing and we whenever we do that simplification, we get that h is equal to 1 .5 g .t .0 squared times 1 .5 g .t .0 minus b0 over g t0 minus b0, and all that is squared.

05:09 So this is going to give us the height.

05:12 For part a, we want to plug in t0 is 1.

05:17 G is 9 .8.

05:21 And h is 20.

05:24 And we want to solve for the v -0 that we have to give the second ball.

05:30 And so whenever we plug these values in here and then solve for v -0, which is best done with the calculator, we get that v -0 is equal to 8 .2 meters per second.

05:44 You're actually going to get another solution for v -0, but that solution is unphysical.

05:49 And so this is the solution that we really care about.

05:52 V .0 is 8 .2 meters per second.

05:59 Now, that might seem like a lot of work just to get the answer to part a, but because we did some heavy algebra and we got some very general expressions here, we're actually going to be able to apply this to our other parts.

06:11 And so for part b, we now want the height whenever v .0 is 6 for the first part.

06:20 So i should label that by an i.

06:22 And so what you do is you take this 6 .0 and you put it into this, and then you're solving for h.

06:28 This time.

06:30 G is still 9 .8 and t .0 is still 1.

06:32 In this case, it's actually easier because our unknown is on the left -hand side, so we just need to calculate the right -hand side with all our nodes...

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