Question

A ball is dropped from a height of 100 m at t = 0. Later, at t = 1.00 s, a second ball is thrown downward with a speed of 19.8 m/s. At what time will the two balls be at the same height?

Answer 1

`88.6 m` above the ground after `1.510 s` of the 1st ball was thrown (`g=10 m/s^2` is taken)

Explanation:

Suppose,after falling through distance `x` from the point of release,both the balls will meet.

Now,for reaching this distance,if the 1st ball takes time,`t`,

Then, `x= 1/2 g*t^2`...`1`

And,for the 2nd ball if `t'` time is required,then,

`x= 19.8t'+1/2g*t'^2`....`2`

Now,given, `t-t'=1` (as `1 s` after the release `1st` ball,the `2nd` ball was thrown)

So,putting the value of `t=1+t'` in equation `1` we can solve the equations `1` &`2;` and we get, `t'=0.510 s`

`And, t=1.510 s`

So,put the value in equation `2` and we get, `x= 11.4 m`

And that means `(100-11.4) m` or `88.6 m` above the ground

Answer 2

`1.49 " seconds"`

Explanation:

`t = sqrt(2*h/g)`
`=> h_1 = t^2*g/2`
`v = v_0 + g t`
`h_2 = 19.8 * (t - 1.00) + g * (t - 1.00)^2 / 2, t >= 1`

`"Here we have :"`
`h_1 = h_2`
`=> t^2*g/2 = 19.8*t - 19.8 + g*t^2/2 - g*t + g/2`
`=> 19.8*t - 19.8 - g*t + g/2 = 0`
`=> (19.8 - g)*t = 19.8 - g/2`
`=> t = (19.8 - g/2)/(19.8 - g)`
`= (19.8 - 4.9)/(19.8 - 9.8)`
`= 14.9/10`
`= 1.49 s`

`g "= gravity constant = 9.8 m/s²"`