A ball is thrown up from the edge of a building. A 2nd ball is dropped from the roof 3s later.........?

Question

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 3 seconds later. If the height of the building is 50 m, what must be the initial velocity of the first ball if both are to hit the ground at the same time?

Answer 1

position = initial position + initial velocity * time + (1/2)*acc.*time^2

Ball thrown up:


y = 50 + v*t - 4.9t^2

Ball dropped down:

0 = 50 - 4.9t^2
t = 3.19 seconds

Since it was dropped 3 seconds after the other ball was thrown upward, the dropped ball took 6.19 seconds to travel to the ground.


At that time, y = 0, since the ball is on the ground
y = 50 + v*t - 4.9t^2
0 = 50 + v*(6.19) - 4.9(6.19)^2
v = [4.9(6.19)^2 - 50] / 6.19
v = 22.25 meters/second

Answer 2

Start by solving the second half of the problem. Knowing the height of the building and the gravity due to acceleration, determine how long it takes the second ball to hit the ground, starting at rest. In a vertical reference frame, y = 0.5*a*t^2. You can derive t for a known y.

Now, take the t you calculated and add 3 seconds to it; I'll call this second time, t'. Then go back and solve the first part of the problem, which is more complicated. You need to use the formula that does not assume zero initial velocity, so y = V_0*t' + 0.5*a*t'^2, where V_0 is the initial velocity, and is the only unknown. Don't forget, V_0 is upwards, but gravity is downwards. The direction from the roof to the ground is also downwards.

This should be enough information for you to complete your homework.

Answer 3

a=d/t^2

The initial velocity of both balls before they are thrown is 0, because they are at rest before you throw them, and they both are subject to an accelaration of: 9,98m/s. (m for meter not mile)
when you either throw or drop them.
If they both hit the ground at the same time the first ball must have been thrown up in the air decelarated by -9.98m/s untill it reached 0 velocity (that's when an object will fall back after it reach higest point) and starts falling from an even higher point.
The distance between the two balls will narrow down untill they both reach the ground.
Now you can evaluate the collision of both balls.
F=ma

Answer 4

a) Calculate how long the 2nd ball take to get to the ground, using 2nd eqn. of motion

b) Add 3 sec. to that time and calculate the distance 1st ball dropped from after it was thrown up (use 2nd eqn. again)

c) Calculate inital velocity of 1st ball using answer from (b) less 50m (use 3rd eqn. of motion, and remember that a is negative)

You should get something less than 60

Sorry.....I'll explain it for you, but I won't do it for you

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Answer 5

a) x = t + a million/2at² 20.1m = t + a million/2(9.81m/s²)t² 40.2m = t + 4.9m/s²t² 0 = -40.2m + t + 4.9m/s²t² 2.77s = t t = t1 + t2 t = 2.77s + a million.08s = 3.85s v / (at) = v0 and x = v0t + a million/2at² positioned them in one yet another x = vt/(at) + a million/2at² x = v/a + a million/2at² 20.1m = v/9.81m/s² + a million/2(9.81m/s²)(3.85s)² 20.1m = v/9.81m/s² + seventy two.70m (-)fifty two.6m= v/9.81m/s² (the minus purely shows course) (-)516.1m²/s² = v (-)22.72m/s = v v = v0 + at -22.72m/s = v0 + (-9.81m/s²)(2.77s) -22.72m/s = v0 - 27.17m/s 4.45m/s = v0 EDIT: I made some blunders alongside the way, i spotted them mutually as i grow to be readin it back, however the way I did it grow to be acceptable. I in simple terms have some random numbers that are no longer stable.

Answer 6

Ask your teacher

Answer 7

Do your own homework and you'll find out.