An object is thrown vertically from a height of $2 m$ $2 m$ at $18 \frac{m}{s}$ $18 m/s$ . How long will it take for the object to hit the ground?

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Answer 1 · 2018-03-19T07:42:52

The time taken for the object to hit the ground is $= 3.78 s$ $=3.78s$

Apply the equation of motion

$s = u t + \frac{1}{2} a t^{2}$ $s=ut+1/2at^2$

The initial velocity is $u = 18 m s^{- 1}$ $u=18ms^-1$

The acceleration due to gravity is $g = 9.8 m s^{- 2}$ $g=9.8ms^-2$

Resolving in the vertical direction ${↑ ⏐ ⏐}^{+}$ $uarr^+$

The distance travelled is $s = - 2 m$ $s=-2m$

Therefore,

$- 2 = 18 t - \frac{1}{2} \cdot 9.8 \cdot t^{2}$ $-2=18t-1/2*9.8*t^2$

Solving this quadratic equation for $t$ $t$

$- 2 = 18 t - 4.9 \cdot t^{2}$ $-2=18t-4.9*t^2$

$4.9 t^{2} - 18 t - 2 = 0$ $4.9t^2-18t-2=0$

$a = 4.9$ $a=4.9$

$b = - 18$ $b=-18$

$c = - 2$ $c=-2$

The discriminant is

$Delta=b^2-4ac=(-18)^2-4(4.9)(-2)=363.2$

Therefore,

$t=(-b+-sqrtDelta)/(2a)=(18+-sqrt(363.2))/(2*4.9)$

$t_1=(18+19.06)/9.8=3.78s$

$t_2=(18-19.06)/9.8=-0.11$

Therefore,

The time taken for the object to hit the ground is $=3.78s$

An object is thrown vertically from a height of 2 m2m2 m at 18 m/s18ms18 m/s. How long will it take for the object to hit the ground?