Question

a ball is thrown up in vacuum returns after 12 s. its position after five seconds will be same as after?

Answer 1

From the equation of displacement in y-direction: 
$y=ut- \frac{1}{2} g t^2$
for t=12s, y=0m, therefore 
$u= \frac{1}{2} gt= \frac{1}{2} \times 10\times 12=60 m/s$
For t=5s , y= 175m.
Substituting this value is the equation of y, we get
$60t- \frac{1}{2}gt^2=175 \\ 5t^2-60t+175 =0$
$t=5s \\ t=7s$
After t=7s it will at the same position as it was at t=5s.

Answer 2

Explanation:

From the equation of displacement in y-direction:

y=ut- \frac{1}{2} g t^2y=ut−

2

1

gt

2

for t=12s, y=0m, therefore

u= \frac{1}{2} gt= \frac{1}{2} \times 10\times 12=60 m/su=

2

1

gt=

2

1

×10×12=60m/s

For t=5s , y= 175m.

Substituting this value is the equation of y, we get

\begin{gathered}60t- \frac{1}{2}gt^2=175 \\ 5t^2-60t+175 =0 \end{gathered}

60t−

2

1

gt

2

=175

5t

2

−60t+175=0

\begin{gathered}t=5s \\ t=7s\end{gathered}

t=5s

t=7s

so answer is 7 sec