Answer:
[tex](175/3)\; {\rm m}[/tex] above the ground, [tex](7/3)[/tex] seconds after the first ball was thrown upward. Assumption: [tex]g = 10\; {\rm m\cdot s^{-2}}[/tex], and that air resistance is negligible.
Explanation:
Under the assumptions, both balls would accelerate downward in the vertical direction at a constant [tex]a = (-g) = (-10)\; {\rm m\cdot s^{-2}}[/tex]. The following SUVAT equation gives the position of one such object:
[tex]\displaystyle x = \frac{1}{2}\, a\, t^{2} + u\, t + x_{0}[/tex],
Where:
Using this equation, obtain an expression for the height of each ball [tex]t[/tex] seconds after the first ball was launched ([tex]t \ge 1[/tex], height is relative to the ground level.)
The initial velocity of the first ball would be [tex]u = 20\; {\rm m\cdot s^{-1}}[/tex] (positive since this velocity points upward.) This ball is launched from ground level, such that [tex]x_{0} = 0\; {\rm m}[/tex]. At [tex]t[/tex] seconds after the first ball was launched, the height of this ball (in meters) would be:
[tex]\displaystyle \frac{1}{2}\, (-10)\, t^{2} + 20\, t[/tex].
Similarly, the initial velocity of the second ball would be [tex]u = (-5)\; {\rm m\cdot s^{-1}}[/tex] (negative since this velocity points downward.) Initial position would be [tex]x_{0} = 35\; {\rm m}[/tex]. However, because this ball would launched [tex]1\; {\rm s}[/tex] after the first ball was launched, all references to [tex]t[/tex] ([tex]t \ge 1[/tex]) in the SUVAT equation should be replaced with [tex](t - 1)[/tex]:
[tex]\displaystyle \frac{1}{2}\, (-10)\, (t - 1)^{2} + (-5)\, (t - 1) + 35[/tex].
Equate the two expressions for height and solve for time [tex]t[/tex] to find the time when the two balls meet:
[tex]\displaystyle \frac{1}{2}\, (-10)\, t^{2} + 20\, t = \frac{1}{2}\, (-10)\, (t - 1)^{2} + (-5)\, (t - 1) + 35[/tex].
[tex]-5\, t^{2} + 20\, t = -5\, t^{2} + 5\, t + 35[/tex].
[tex]\displaystyle t = \frac{7}{3}[/tex].
In other words, the two balls would meet at [tex](7/3)[/tex] seconds after the first ball was launched.
Substitute the value of [tex]t[/tex] into either expression to find the height of the two balls when they meet:
[tex]\begin{aligned}& \frac{1}{2}\, (-10)\, t^{2} + 20\, t \\ =\; & \frac{1}{2}\, (-10)\, \left(\frac{7}{3}\right)^{2} + 20\, \left(\frac{7}{3}\right) \\ =\; & \frac{175}{9}\end{aligned}[/tex].
In other words, the two balls would meet at a height of [tex](175/9)[/tex] meters relative to the ground level.
