Question

Two balls are thrown vertically upward, with the same initial velocity of $98{\text{m s}}^{-1}$. The second ball is thrown $4\text{s}$ after the first one. How long, after the first one is thrown, will they meet? $(g=9.8{\text{ms}}^2)$

• A. $t=22\text{s}$
• B. $t=8\text{s}$
• C. $t=12\text{s}$
• D. $t=16\text{s}$

Answer 1

The correct option is C $t=12\text{s}$

Answer (2)
Displacement of both balls must be same at the instant they meet
$x_1=x_2$
$98t-\frac{1}{2}g{t}^2=98(t-4)-\frac{1}{2}g{(t-4)}^2$
$98t-\frac{1}{2}g{t}^2=98t-98\times 4-\frac{1}{2}g{t}^2-\frac{1}{2}g\times 16+\frac{1}{2}g\times 8t$
$98\times 4=\frac{1}{2}g(8t-16)$
On solving, $t=12$ seconds