A rock is thrown from the top of a tall building. Then distance, in feet, between the rock and the ground t seconds after it is thrown is given by d=-16t^2-4t+382. After how many seconds wil it be at a distance of 370 feet from the ground?

1 Answer
Jun 1, 2018

Equation of motion of a rock thrown from the top of a tall building is

#d=-16t^2-4t+382#

(Compare it with kinematic expression

#h=h_0+ut-1/2 g t^2#)

Time #t_1# when it is #370\ feet# from ground is

#370=-16t_1^2-4t_1+382#
#=>16t_1^2+4t_1-12=0#
#=>4t_1^2+t_1-3=0#

Solving using split the middle term we get

#4t_1^2+4t_1-3t_1-3=0#
#=>4t_1(t_1+1)-3(t_1+1)=0#
#=>(4t_1-3)(t_1+1)=0#
Roots are #(4t_1-3)=0and(t_1+1)=0#
which gives us #t_1=3/4 and -1\ s#

Since time can not be negative we have

#t_1=3/4\ s#