Question

At $t=0$, an arrow is fired vertically upwards with a speed of $98m{s}^{-1}$. A second arrow is fired vertically upwards with the same speed at $t=5s$. Then, (Take $g=9.8m{s}^{-2}$)

• A. the arrows will be at the same height above the ground at $t=12.5s$.
• B. the two arrows will reach back at their starting points at $t=20s$ and $t=25s$.
• C. the ratio of the speeds of the first and the second arrows at $t=20s$ will be $2:1$.
• D. all are correct.

Answer 1

Answer: (D)

all are correct.

$s=ut+\frac{1}{2}a{t}^2\Rightarrow 98t-\frac{1}{2}9.8t^2=98(t-5)-\frac{1}{2}9.8{(t-5)}^2\Rightarrow t^2-{(t-5)}^2-100=0\Rightarrow 10t=125t=12.5s$
Now, total time for first arrow will be
$s=ut+\frac{1}{2}a{t}^2\Rightarrow 98t-\frac{1}{2}9.8t^2=0t(t-20)=0t_1=20s\Rightarrow t_2=25s$
Now at $t=20s$
$v=u+at{v}_1=98\&v_2=98\frac{1}{2}\frac{v_1}{v_2}=\frac{2}{1}$