A hawk flying at a height of 50 feet spots a mouse on the gr

A hawk, flying at a height of 50 feet, spots a mouse on the ground. He dives to catch the mouse at a speed of 45 feet per second. a. Using the vertical motion model, h(t) = -16t^2 +vt + h_o, write the function of the flight of the hawk in respect to time. a. Assuming that the rat is stationary, how long does it take the hawk to catch the mouse? 9. You are competing in high jump. You need to jump 6 feet in the air to clear the bar. The height of your body above the ground is given by the function: h(t) = -16t^2 +20t. What is the maximum height your body will be above the ground? Explain whether or not you are able to clear the bar.

Solution

8) Model equation : h(t) = -16t^2 +vt +h

h ---intial height , v =velocity

given : h = 50 ft ; v = 45 f/sec

plug these values:

h(t) = -16t^2 + 50 +45t

a) when hawk reaches ground he catches mouse i.e. h(t) =0

-16t^2 + 50 +45t =0

solve for time t : use quadratice root formula:

ax^2 +bx +c =0

x = ( -b +/- sqrt(b^2 -4ac) )/2a

=( -45 +/- sqrt(45^2 +4*16*50 )/-32

x = -0.852 , x = 3.67 neglect -ve root

so, x = 3.67 sec hawk takes to catch mouse on ground

9) h(t) = -16t^2 +20t

maximum height is given by the vertex of a quadratice function : ax^2 +bx +c =0

vertex (h, k) ; h = -b/2a = -20/(-2*16) = 5/8 sec = 0.625 sec

maximum height h(0.625) = -16*(0.625)^2 +20*0.625 = 6.25 feet

Yes we are able to clear the bar which is 6 feet only