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Help Solve This Physics Question by uzziwise: 7:57pm On Jan 30, 2020
A ball thrown upward with speed 12m/s from top of a building how much later must a Second ball be dropped from the same starting point if it is to hit the ground at the same time as first ball? The initial position is 24m above the ground.?
Ans:1.25secs
This was how the question was presented from question 15 in the pic below

Re: Help Solve This Physics Question by Mario1989(m): 7:59pm On Jan 30, 2020
Ayam Cuming... Lemme call ALBERT EINSTEIN from the other side to help us...
Re: Help Solve This Physics Question by healthserve(m): 8:02pm On Jan 30, 2020
Consider variables and constant when puzzling the solution to such questions.

Variable here is Time

Constant here is Height and Speed. I.e Velocity and Distance. Slot it into the equation and you'll have the working formula to solving such questions. I.e since Speed = Distance x Time. Time = Speed/Distance. Get the variables for both balls slot it into the equation and you'll have your answer


I hope it's clear
Re: Help Solve This Physics Question by MickzyDonald: 8:34pm On Jan 30, 2020
Bro �, I'm about to solve the equation when the ball i threw landed on an old woman's head
Now I'm in serious trouble

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Re: Help Solve This Physics Question by healthserve(m): 8:35pm On Jan 30, 2020
Shibaraba FranchasNg Pansophist make una come solve physics question o
Re: Help Solve This Physics Question by franchasng: 9:07pm On Jan 30, 2020
healthserve:
Shibaraba FranchasNg Pansophist make una come solve physics question o
The pursuit for paper called money has shattered my small wisdom.


But the little I can remember to assist young students who may want to learn, this problem is a projectile motion problem, and you can find the time it will take for a projectile to reach its maximum height by applying the formula (0 - V) / -32.2 ft/s^2 = T


From the above formula, V is the initial vertical velocity found in step 2. In this formula, 0 represents the vertical velocity of the projectile at its peak and -32.2 ft/s^2 represents the acceleration due to gravity.

If you use this formula you will find the answer shocked
Re: Help Solve This Physics Question by healthserve(m): 9:10pm On Jan 30, 2020
franchasng:
The pursuit for paper called money has shattered my small wisdom.


But the little I can remember to assist young students who may want to learn, this problem is a projectile motion problem, and you can find the time it will take for a projectile to reach its maximum height by applying the formula (0 - V) / -32.2 ft/s^2 = T


From the above formula, V is the initial vertical velocity found in step 2. In this formula, 0 represents the vertical velocity of the projectile at its peak and -32.2 ft/s^2 represents the acceleration due to gravity.

If you use this formula you will find the answer shocked



Me too o. There's a formula specifically for projectiles that utilises (U1 -U2)/2g. Something like this. Omo e don tey o. The thing still dey
Re: Help Solve This Physics Question by franchasng: 9:24pm On Jan 30, 2020
healthserve:




Me too o. There's a formula specifically for projectiles that utilises (U1 -U2)/2g. Something like this. Omo e don tey o. The thing still dey
lol
Re: Help Solve This Physics Question by Nobody: 10:51pm On Jan 30, 2020
healthserve:
baba I Don go take my 33 with pepper soup. My head engine Don knock
Re: Help Solve This Physics Question by Hahjascho(m): 11:51pm On Jan 30, 2020
1. First getting the time it takes the first ball to be thrown upward..using V=u + gt (when an object is thrown upward, u is maximum and v(final veloc. is zero) and g is 10m/s2 (negative when going up.)

Then 0= 12-10t.......t = 1.2sec. (time to move up from the initial point which is the top of the building.

Then use it to get the distance travelled..(H = ut -1/2gt^2.). or rather directly by using

V^2 = U^2 -2gh

12^2 = 0 - 20h .....h = 7.2m....(i.e the distance to move up from the initial point upward. or from up to the initial point ( i.e top of the building).

Total distance (s) = 24m + 7.2 (to reach ground floor).

S = ut + 1/2 gt^2. (moving down, u is 0)


31.2 = ut + 0.5gt^2

31.2 = 0 + 5t^2 t = 2.5sec (I.e the total time to reach the ground..)

So, time for second ball to reach the ground will now be 2.5-1.2 = 1.3 sec.

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Re: Help Solve This Physics Question by Martinez39s(m): 12:32am On Jan 31, 2020
uzziwise:
A ball thrown upward with speed 12m/s from top of a building how much later must a Second ball be dropped from the same starting point if it is to hit the ground at the same time as first ball? The initial position is 24m above the ground.?
Ans:1.25secs
This was how the question was presented from question 15 in the pic below
The answer is 3.5179 sec. Should I explain?
Re: Help Solve This Physics Question by uzziwise: 8:25am On Jan 31, 2020
Yes please
Martinez39s:
The answer is 3.5179 sec. Should I explain?
Re: Help Solve This Physics Question by uzziwise: 8:26am On Jan 31, 2020
I really appreciate all your contributions . The have all been helpful.
Re: Help Solve This Physics Question by Martinez39s(m): 9:13am On Jan 31, 2020
uzziwise:
I really appreciate all your contributions . The have all been helpful.
The position function (the function that gives the distance of the object from the ground in terms of time) of the first stone is
s(t) = - 4.9t² - 12t + 24 ———(1)
t ≥ 0

The position function of the second stone is
s(ţ) = - 4.9ţ² + 24 ——— (2)
ţ ≥ 0
[/quote]

We have that
ţ = t - c ——— (3)
where c is the time required to start throwing the second stone so that it simultaneously reaches the ground with the first stone.


YOUR TASK :
(A) solve (1) when s(t) = 0.
(B) Use (3) to write (2) in terms of the t. The new function [ s(t - c) = - 4.9(t - c)² + 24 ] is equal to 0 when t is equal to the solution gotten from (A) (since both stone reached the ground at the same time).
(C) Solve for c and get your answer.


Any question?

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