John hit a golf ball from the ground and it followed the projectile h(t)=-16t²+100t+0, where is the time in seconds, and ℎ is the height of the ball. Use the quadratic formula to solve the two roots. These represent the 2 times that the ball is on the ground. At what time does the golf ball land? What is the max height of the golf ball? At what time (t) is the golf ball at the maximum height?

The maximum height of the golf ball h(t) = 156.25m

The maximum height of the golf ball at  t = 3.125 seconds

Step-by-step explanation:

Step(i):-

Given

                 h(t)=-16t²+100t   ...(i)

Differentiating equation (i) with respective to 't'

              -32t + 100 =0

            ⇒ -32 t = -100

         t = 3.125 > 0

Step(ii):-

 Given h(t) = -16t²+100t  = 0

t ( -16 t + 100 ) =0

t =0 and  -16t = -100

t =0 and  t = 6.25

Step(iii):-

             h(t) = -16t²+100t ...(i)

put t=0  in equation (i)

           h(0) = 0

put t = 3.125 in equation (i)

h(t) = -16(3.125)²+100(3.125)

         h (t) =  156.25

Put t = 6.25

h(t) = -16(6.25)²+100(6.25)

    = 0

The maximum value h(t) = 156.25 at 3.125

final answer:-

The maximum height of the golf ball h(t) = 156.25m

The maximum height of the golf ball at  t = 3.125 seconds

t = 3

step-by-step explanation:

put t = 3 in this expression :

(3²-1)(3²-2)(3² - 3)(3² - 4) = 8×7×6×5 = 56×30 = 1680

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.