Free Fall Problem?

Question

A person is standing on the edge of a cliff. He throws a ball vertically up at 8.8mls. After a delay of 1.0s, he drops a second ball. Assume both balls accelerate uniformly at 9.8m/s2 [down]. What distance has been travelled by the second ball when the first ball catches up to it?

So i know initial velocity = 8.8 m/s and acceleration is 9.8m/s2, but what else do i have to find before i can solve this question?

Answer 1

Distance moved by the first ball in T second is
S = uT + 0.5 a T^2

S = - 8.8 T + 4.9 TT (since 0.5 g = 4.9)

Distance moved by the second ball in (T - 1) second is the same as S.

S = 4.9 (T-1)^2.

Therefore,

- 8.8 T + 4.9 T^2 = 4.9 (T-1)^2.

Solving we get T = 4.9 second.

using S = 4.9 (T-1)^2 or S = - 8.8 T + 4.9 TT

S = 74.529 m.

Answer 2

You don't need any more information to solve this problem. It might make it easier, but it isn't necessary
I will set up a coordinate plan with y = 0 as the height of the cliff, so positive values mean above the cliff, and negative values mean below the cliff. The equation for motion of the first ball is y1 = 8.8t – ½ at², so at t = 1, y =3.9m, so the first ball is 3.9m above the cliff when the second ball is released. Using v = v0 + at, we find that the first ball has a speed of -1m/s, the negative showing that the velocity is directed downward.
The next part is a bit tricky. If we use the equation of motion for the second ball we get y2 = - ½ at², but this cannot be used with equation 1 to solve for the time, if you try you will get the answer t=0. This isn’t wrong, at t=0 both balls are at the same height, which is what we are solving for, but this isn’t what we would expect. The reason why is because we can’t equate two equations that don’t refer to the same timeline. At t=1 in equation one we find that the height of the ball is 3.9m above the cliff, but in equation 2 we find the position of the second ball is 4.9m below the cliff, it should be 0. This discrepancy is caused by the one second delay between the start of flight, this delay must be accounted for. The easiest way to do this is to start the timeline from when the second ball is dropped, at this time v0 of the first ball is -1m/s, this value must be used in the new equation for ball 1. We must also add 3.9 to the first equation, because the ball is 3.9m above the cliff when the second ball is dropped. That explanation concluded, we have y1 = -t – ½ at² +3.9m, and y2 = - ½ at², solving these yields t = 3.9s as the intersection of these two functions.
Now, as you can see most of my answer had been taken up with solving for the time, this might seem strange, as you asked for the distance the first ball traveled before the it reached the second ball. We need this time, however, to compute the distance. You did ask for the distance and not the displacement, which would be much easier to find, so don’t blame me if the answer gets complicated.
The distance is the same as the arc length of the displacement function. The formula for the arc length of any function is ∫ab (1+f’(x)²)1/2 dx. The limits a and b are the beginning and end of the first ball’s flight, in our case a=0 and b=3.9. f’(x) is the derivative of the motion equation, so in our case f’(x) = -1-9.8x. These types of integrals can be tough to solve so I used a numerical solution and obtained 78.61m as an approximation. If you don't have a numerical solver, just take a minute to construct to create a Taylor series. Just for your reference, the displacement is -74.53m. I assume that neither of the balls will hit the ground before they intersect.
I hope this helps.

Answer 3

You need to know where the first ball is relative to the release point of the second ball; and you need to know which direction the first ball is going at the time of the release of the second ball. You can use the SUVAT equations [See source.] for all these calculations...starting with t = 1 sec, u = 8.8 m/sec, and g = 9.8 m/sec^2.

To find which direction ball one is going when t = 1 sec, just plug in v = u + gt; where v = 0 at the apex of ball one's upward travel, u = 8.8 m/sec initial upward velocity, and g = 9.8 m/sec^2. Solve for t = u/g = 8.8/9.8 < 1 sec; so the ball has reached the top of its travels and is about .1 sec into its downward trek when ball two is dropped.

The distance of the apex above release point at t ~ .9 sec is found by h = (u/2)t = 4.4 X .9 ~ 4 meters. u/2 is the average velocity during its upward travel = (u + v)/2; where v = 0 at the apex. So in another t = .1 sec, the first ball will have fallen s = 1/2 gt^2 = 1/2 9.8 .01 ~ .05 meters. Now we can reset the clock to t = 0 at the time the second ball is dropped.

At t = 0 the two balls are about d = 4 - .05 = 3.95 meters apart. Ball one is traveling at v(1) = gt = 9.8 X .1 ~ 1 m/sec. Ball two is traveling at v(2) = 0 because it was just released. Thus the rate of closure (r = relative velocity) is 1 m/sec between the two balls. Since they started 3.95 meters apart at the reset t = 0, it will take t = d/r = 3.95/1 ~ 3.95 sec before ball one catches up with ball two.

We can treat r = 1 m/sec = constant because both balls are increasing their velocities at the same rate (g = 9.8 m/sec^2). Thus, v(2) - v(1) = u(1) + gt - u(2) - gt = u(1) - u(2) = (1 - 0) = 1 m/sec = r; we see that the accelerating terms cancel out since they are both gt.

In t = 3.95 sec, measured from the time ball two is dropped, ball two will have dropped s = 1/2 gt^2 = 4.9 X 16 ~ 80 meters or thereabouts.

Pardon the approximations, I figure you want more to know how the problem is worked rather than the precise answer.

Lesson learned: The tricky part is to know where the two balls are in relation to each other at the time the second ball is released. For example, had the second ball been dropped at, say, t = .5 sec, the first ball would still be traveling upward and widening the distance (d) between them. Also note, once the relative closure rate (relative velocity) was found, acceleration was no longer a factor because both balls are accelerating at the same rate.

Answer 4

The 9.8 m/s/s acceleration of gravity is down, so it should be a negative number, -9.8. The initial velocity is up, so it should be a positive number +8.8. After one second it is moving at 8.8-9.8=-1.0 m/s, where the - sigin means down, but it is still at some height above where you threw it from. At that moment the other ball is dropped from zero height starting at zero velocity with -9.8 m/s/s acceleration. First you need to calculate the height of the first ball at the one second time, then if it is X meters, then it will catch up with the other ball in x seconds because both balls are accelerating at the same rate, but one started out 1 m/s faster than the other. In other words, the first ball is always falling 1 m/s faster than the second ball. After you know the number of seconds, you can calculate how far the second ball falls in that time, and that is the place where the first ball catches up. Useful formulas are:
X=V0*T+0.5*A*T^2
V=V0*T+A*T
X^2=V^2+2*A*X

Answer 5

Ok, so you know initial velocity of the first ball is 8.8m/s.

With s=V0+1/2at^2, you can see that the displacement of the first ball after 1.0s is 3.9m.

Now, if the person DROPS the second ball, instead of throwing it up, then the second ball is going to maintain it's 3.9m below the first ball, because they both accelerate at -9.8m/s^2. Thus, the first ball will not catch up to the second ball until the second one hits the ground, and bounces, and crosses the same path in the y direction as the first ball. Or, in short, basically until they both come to rest on the ground.

Answer 6

y1(t) = v*t - g*t^2/2 - the equation of motion for the first ball

y2(t) = -g*(t - t0)^2/2 - the equation of motion for the second ball

where v = 8.8 m/s, t0 = 1.0 s

y1(t1) = y2(t1) - the first ball catches up with the second ball at

the time t1 (counted from the throwing of the first ball)
We get a linear equation in t1:
v*t1 - g*t1^2/2 = - g*(t1-t0)^2/2

2*v*t1 - g*t1^2 = -g*t1^2 + 2*g*t0*t1 - g*t0^2

2*(v - g*t0)*t1 = -g*t0^2

t1 = g*t0^2/(2*(g*t0 - v))

Please note that if g*t0 < v, then there is no solution to the problem! In our case 9.8 m/s^2 * 1.0 s = 9.8 m/s > 8.8 m/s, so we have a solution.

t1 = 9.8 m/s^2 x (1.0 s)^2/(2 x (9.8 m/s^2 x 1.0 - 8.8 m/s)) =
= 9.8/2.0 s = 4.9 s

The coordinates of the balls are:
y1 = y2 = -9.8 m/s^2 x (4.9 s - 1.0 s)^2 / 2 = -150 m

The distance traveled by the second ball is equal to its displacement, because it moves in a straight line and it doesn't change its direction (like the first ball), so:

s2 = |y2 - 0| = |y2| = 150 m

Answer 7

there is definitely something missing

you need either the maximum height reached by the first ball or the height of the cliff or the acceleration with which the first ball is projected.

Answer 8

You, personaly, don't have to find out anything else...whoever came up with this problem should have added the hight of the cliff or the amount of time it would take for one of he balls to reach the ground ( and given the speed and acceleration you then would have calculated the hight of the cliff). In this state, teh problem can not be solved...if your teacher gave you this problem ask him/her about any potential missing information... if not, the try to solve it with a fictional value given by yourself (e.g. hight of cliff= 120 ft, or time neede for second ball to reach ground=40sec)...
Note: the numbers i gave you in the last 2 rows are purely fictional.

Answer 9

at the same time as he weighs four hundred N, the elevator is accelerating down. at the same time as he weighs 650 N, contained in the 2d 1/2 of the mission, the elevator is accelerating up. contained in the first section, he "loses" a million/3 of his weight to the acceleration of the elevator, so x / 9.8 = a million / 3 x = 3.27 m/s^2 For the 2d section, the physics student effectual factors a million/12 of his weight with the aid of acceleration, so x / 9.8 = a million / 12

Answer 10

I think u need the height of cliff, or time taken.. of U,V,A,S,T u only had 3