If an object is dropped, how fast will it be moving after falling $16 m$ $16 m$ ?

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Answer 1 · 2017-01-12T14:59:25

The velocity is $= 17.7 m s^{- 1}$ $=17.7ms^-1$

Computing in the vertical direction ${⏐ ↓}_{+}$ $darr_+$

$u = 0$ $u=0$

$a = g = 9.8$ $a=g=9.8$

$s = 16$ $s=16$

$v^{2} = u^{2} + 2 a s$ $v^2=u^2+2as$

$v^{2} = 0 + 2 \cdot 9.8 \cdot 16$ $v^2=0+2*9.8*16$

$v = \sqrt{2 \cdot 9.8 \cdot 16} = 17.7 m s^{- 1}$ $v=sqrt(2*9.8*16)=17.7ms^-1$

Answer 2 · 2017-01-12T15:01:35

$v = \frac{28 \sqrt{10}}{5} m s^{- 1}$ $v=(28sqrt10)/5ms^(-1)$

measuring down as positive

Considering vertical motion

initial velocity $u = 0 m s^{- 1}$ $u=0 ms^(-1)$

acceleration $g = + 9.8 m s^{_2}$ $g=+9.8ms^(_2)$

displacement $s = 16 m$ $s=16m$

$v^{2} = u^{2} + 2 a s$ $v^2=u^2+2as$

$v^{2} = 0 + 2 \times 9.8 \times 16$ $v^2=0+2xx9.8xx16$

$v^{2} = 313.6$ $v^2=313.6$

$v = \sqrt{313.6} = \frac{28 \sqrt{10}}{5} m s^{- 1}$ $v=sqrt313.6=(28sqrt10)/5ms^(-1)$

If an object is dropped, how fast will it be moving after falling 16m16 m?