Question

can u plzz solve thia ques?​

Answer 1

$\Huge{\underline{\underline{\blue{\mathfrak{\star\ Solution\ \star}}}}}$

Initial velocity u = 40 m/s

Acceleration a = -g = -10 m/s²

Final velocity at the highest point would be v = 0

Let the maximum height reached be s

As per the third equation of motion

$\sf {v^{2}-u^{2} = 2as}\\\\\sf {s = \frac {v^{2}-u^{2}}{2a}}\\\\\sf {s = \frac {0^{2}-40^{2}}{2\times -10}}\\\\\sf {s=80m}$

The net displacement would be zero as the stone will return to the point from where it was thrown.

The total distance covered by the stone = 2s = 160 m

Answer 2

Answer

\Huge{\underline{\underline{\blue{\mathfrak{\star\ Solution\ \star}}}}}

Initial velocity u = 40 m/s

Acceleration a = -g = -10 m/s²

Final velocity at the highest point would be v = 0

Let the maximum height reached be s

As per the third equation of motion

\begin{gathered}\sf {v^{2}-u^{2} = 2as}\\\\\sf {s = \frac {v^{2}-u^{2}}{2a}}\\\\\sf {s = \frac {0^{2}-40^{2}}{2\times -10}}\\\\\sf {s=80m}\end{gathered}

The net displacement would be zero as the stone will return to the point from where it was thrown.

The total distance covered by the stone = 2s = 160 m