same initial speed & both hit the ground below the cliff. which ball hits the ground at greater speed? ignore air resistance.
this is for an algebra based phys class.
2006-10-09 09:53:22 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The one thrown up. Gravity accelerates both balls at 9.8 m/s^2. The one going up has more time to accelerate -going down -before hitting the ground. You're welcome : D.
2006-10-09 09:58:09 · answer #1 · answered by Chuglon 3 · 1⤊ 5⤋
does gravity take effect? If so - A body in motion stays in motion. Its result would be that the one thrown up in the air would gain more momentum on it's way down and hit the ground with greater speed, but pretty much land at the same time (Unless the cliff is only 10 feet down - I am assuming a drop of at least 100 feet or more.) as the one thrown straight down. If it was a short cliff the object thrown directly down would hit first but with less impact as the one thrown straight up.
2006-10-09 10:01:48 · answer #2 · answered by Anonymous · 0⤊ 2⤋
This formula will allow you to find the final velocity, given the initial velocity and the height from the ground:
v^2 = vo^2 + 2adY
I will now solve for v by taking the square root:
v = sqrt(vo^2 + 2adY)
Since the initial velocity is the same in both cases, the only difference between throwing the ball up and throwing the ball down is the direction in which it's being thrown. Throwing a ball up will give a positive velocity, and throwing it down will give a negative velocity. BUT, as you can see from the rearranged formula, it doesn't matter if the initial velocity (vo) is negative or positive, because it is being squared in the formula. Because it's being squared, as long as the magnitude of th velocity is the same, it doesn't matter if the one ball is thrown up and one is thrown down; both balls will always have the same velocity as they hit the ground.
2006-10-09 10:08:03 · answer #3 · answered by عبد الله (ドラゴン) 5 · 0⤊ 1⤋
They both hit the ground with the same speed!
The ball thrown upward will rise, stop, and fall back through the starting point with the same speed it was released. As stated, this will be the same speed the ball thrown downward was released with.
2006-10-09 10:18:50 · answer #4 · answered by entropy 3 · 4⤊ 0⤋
a) Plug in t=o by way of fact the ball has not coated any distance. Its at time 0 and we ought to how extreme the development is so as that provides you one hundred forty four b) Now while the ball hits the floor, the h=0. so positioned s=0 and sparkling up for t fee, likely you get 2 values. +ve one ought to be the respond on the grounds that time is in no way detrimental c) optimal top. Take the spinoff of the equation s' = -32t + ninety six and now positioned s'=0. whenever you seem for maximum/min, continuously positioned your spinoff = 0 0 = -32t + ninety six t = ninety six/32 t = 3 at t = 3 the ball reaches the max top and to discover what the max top is, plug t = 3 into the unique equation s = -16t^2 + 96t + one hundred forty four wish this facilitates :)
2016-11-27 03:13:18 · answer #5 · answered by Anonymous · 0⤊ 0⤋
Both balls hit with the same speed. The ball thrown up decelerates as it goes up and then begins to accelerate as it comes down, attaining the same speed as it passes the thrower that it was thrown, it then continues to accelerate to match the ball that was thrown down.
2006-10-09 10:06:14 · answer #6 · answered by green star 3 · 2⤊ 0⤋
my first thought is "how hard is the ball thrown"... but since that info isn't provided:
both hit the ground at the same speed
2006-10-09 09:57:05 · answer #7 · answered by bequalming 5 · 1⤊ 0⤋
If the cliff is sufficiently tall for the balls to reach terminal velocity, they will hit at the same speed.
2006-10-09 10:01:48 · answer #8 · answered by Anonymous · 1⤊ 2⤋
Do your own homework!
2006-10-09 09:56:14 · answer #9 · answered by Zebra4 5 · 1⤊ 2⤋