a person standing near the edge of a cliff 100ft above a lak

a person standing near the edge of a cliff 100ft above a lake throws a rock upward with an initial speed of 32 feet per second. The height of the rock above the lake at the bottom of the cliff is a function of time and is described by h(t)=-16t+32t+100, how many seconds will it take until the rock reaches maximum height and what is the height and at what time will the rock hit the water

Solution

given height of the rock above the lake at the bottom of the cliff is a function of time and is described by  h(t)=-16t²+32t+100

for maximum height dh/dt =0

=>-16*2t+32*1+0=0

=>-32t=-32

=>t=1 sec

it takes 1 sec to reach maximum height for the rock

maximum height =h(1)=-16(1)²+32(1)+100

maximum height =h(1)=116ft

when rock hits the water h(t)=0

-16t²+32t+100=0

16t²-32t-100=0

4t²-8t-25=0

roots of ax^2 +bx +c =0 ==>x=[-b+(b²-4ac)]/(2a),x=[-b-(b²-4ac)]/(2a)

a =4 , b=-8 ,c=-25

==>t=[8+(64+400)]/(2*4),t=[8-(64+400)]/(2*4)

==>t=3.69sec ,t=-1.69

t cannot be negative

so it takes 3.69 sec for the rock to hit the water