Algebra 2 BIM Chapter 2 Quiz 2

Question

A hockey player strikes a hockey puck. The height of the puck increases until it reaches a maximum height of 1.5 feet, 30 feet away from the player. The height  (in feet) of a second hockey puck is modeled by , where  is the horizontal distance (in feet). Compare the distances traveled by the hockey pucks before hitting the ground.

Answer 1

gravity is the same for every puck. initial height is the same, zero

only difference is initial velocity, how hard the puck was struck

general equation is -16t^2 + vt = h(t) = 1.5 at maximum height, height as a function of time

take the first derivative and set it equal to zero for the time at maximum height = 1.5 feet

-32t + v = 0

multiply by t to get

-32t^2 + vt = 0

subtract this from the general equation to get

16t^2 = 1.5

divide by 16 to get

t^2 = 1.5/16

take the square root of both sides

t= sqr1.5/16 = 1/4sqr1.5=(1/4)x 1.22 seconds at time of maximum height and distance = 30 feet

divide 1.22 by 4 to get .305 seconds, round up .31 seconds almost a 1/3 of a second. If you've seen a hockey game, that sounds about right.

then v = 32(.31) = 9.92 feet per second as the initial velocity

IF a puck is hit harder, it goes further than 2(30)= 60 feet, hit with less initial velocity and it goes less than 60 feet. The puck goes twice the distance it goes when it reaches its maximum height.

Answer 2

Compare the distances traveled by the hockey pucks before hitting the ground.

This sounds like you want side "c".

As for side "c"; a^2+b^2=c^2.

1.5^2+30^2=c^2

2.25+900=902.25^.5=30.03747.............ft

Triangle: side "a" shortest

b mid

c hypotenuse