Asoccer player kicks a ball with an initial velocity of 35 feet per second from a starting point of 0 feet. use the equation h(t)=-16t^2+bt+c, where v is the initial velocity in feet/second and c is the height in feet. round to the nearest tenth. #1 calculate how long the ball was in the air #2 calculate the maximum height of the ballasap

2.2 seconds

Step-by-step explanation:

h(t)=-16t^2+bt+c

We know the initial velocity is 35  = b

We know the initial height is 0 =c

h(t)=-16t^2+35t+0

h(t)=-16t^2+35t

Question 1 is to find when the ball hits the ground  or when h(t) is 0

0=-16t^2+35t

Factor out a t

0 = t( -16t+35)

We can use the zero product property  to solve for t

t=0   -16t+35 =0

          Subtract 35 from each side

         -16t = -35

          Divide by -16

          -16t/-16  = -35/-16

                t = 2.1875 seconds

t=0 is when the ball is kicked

t =2.1875 is when the ball lands on the ground after being kicked

Rounded to the nearest tenth

2.2 seconds

19.1 ft

The maximum height is at the axis of symmetry

h = -b/2a

  -35/(2*-16)

  -35/-32

  35/32

This is the t value

To find the maximum height , we need to put this in the equation to find the y value

h(35/32) = -16( 35/32)^2 + 35 *(35/32)

             = -16* 1.196289 +38.28125

               19.140625 ft

Rounding to the nearest tenth

19.1 ft

right

step-by-step explanation:

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.