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
#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#
