An object is thrown vertically from a height of $2 m$ $2 m$ at $21 \frac{m}{s}$ $21 m/s$ . How long will it take for the object to hit the ground?

Question

1586 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-06T02:21:13

$0.092 s$ $0.092 s$

Let's find the velocity with which it will hit the ground,applying

$v^{2} = u^{2} + 2 g s$ $v^2=u^2 +2gs$

Given, $u = 21 m s^{- 1}, s = 2 m$ $u=21ms^-1,s=2m$

So, $v^{2} = 21 \times 21 + 2 \times 9.8 \times 2$ $v^2=21×21 +2×9.8×2$

So, $v = 21.92 m s^{- 1}$ $v=21.92ms^-1$

Now,if it took time $t$ $t$ in gaining that much velocity we can use, $v = u + g t$ $v=u+g t$

So, $21.92 = 21 + 9.8 t$ $21.92=21 + 9.8t$

Or, $t = 0.092 s$ $t=0.092 s$

An object is thrown vertically from a height of 2m2 m at 21ms21 m/s. How long will it take for the object to hit the ground?