An object is thrown vertically from a height of $2 m$ at $18 m/s$ . 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 $=3.78s$

Apply the equation of motion

$s=ut+1/2at^2$

The initial velocity is $u=18ms^-1$

The acceleration due to gravity is $g=9.8ms^-2$

Resolving in the vertical direction $uarr^+$

The distance travelled is $s=-2m$

Therefore,

$-2=18t-1/2*9.8*t^2$

Solving this quadratic equation for $t$

$-2=18t-4.9*t^2$

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

$a=4.9$

$b=-18$

$c=-2$

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 m2 m at 18 m/s18 m/s. How long will it take for the object to hit the ground?