Question

A ball is dropped from the top of a building. One second later, another ball is thrown from the top of the building with a velocity of 30m/s. Will the balls ever meet? If so, when and where?

Answer 1

Any attempt ?

Answer 2

none yet. Lack of given?

Answer 3

Info is sufficient .

Answer 4

You just need to use newton;s 2nd eqn

Answer 5

we have to use formulas.

Answer 6

Same thing :|

Answer 7

how?

Answer 8

i assume you mean 'meet in the air'.
it depends on the height of the building,

the formula for the distance travelled is s = ut + 0.5gt^2
so in 1 second the ball that is just dropped travels 0(1) + 16(1)^2 = 16 ft
so if the building is equal or less than 16ft high they will not meet in the air.

Answer 9

otherwise cosier the flight of the 2 balls:

for first ball travelling for t seconds

s = 16t^2

for the second ball:

suppose they meet in the air then the second ball has travelled the same distance s and has been in the air for (t - 1) seconds) so

s = 30(t - 1)+ 16(t - 1)^2

so
30(t - 1)+ 16(t - 1)^2 = 16t^2

30t - 30 + 16t^2 - 16t^2 - 32t + 16 = 0
-2t = 14
t = -7

which is impossible

so they never meet

Answer 10

does that make sense amistre? is my method valid?

Answer 11

im contemplating it :) id use 9.8 as gravity just to be sure since the velocity given is in metters

Answer 12

the dropped ball has +1 seconds on it; while the throwm ball has +0 seconds to cover the same distance.

Answer 13

g(t+1)^2/2 = gt^2/2 + vt

g(t+1)^2 = gt^2 + 2vt

gt^2 +2gt +g = gt^2 + 2vt

2gt +g = 2vt

2gt - 2vt = -g

t(2g - 2v) = -g

t = -g/(2g - 2v)

t = g/(2v - 2g)

Answer 14

9.8/2(30-9.8) is about .25 meters

Answer 15

err, seconds lol

Answer 16

if we use a gravity of 32 feet per second on this then its negative .. but then the dimensions dont match up

Answer 17

silly me - i used 30 ft/sec instead of 30 m/sec

Answer 18

or even 9.8 m/sec :)

Answer 19

if you replace 32 ft/s by 9.8 m/s and repeat my work the answer comes to 1.24 seconds that first ball is in the air and the distance where they meet = 7.53 metres