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#