Question

A golf ball is projected upward from ground level at an initial velocity of 112 ft/sec. The height of a projectile can be modeled by s(t) = – 16t^2 + v_0t + s_0, where t is time in seconds, s_0 is the initial height in feet, and v_0 is the initial velocity in ft/sec.

Answer 1

The ball reaches a maximum height of 196ft above the point where it was launched.

Step-by-step explanation:

Please see attachment below.

Answer 2

maximum height: 196

height of ball after 2 seconds: 160

ball reaches ground: 7 seconds

Step-by-step explanation:

height of ball after 2 seconds:

s(t)=-16t^2+112t

s(2)= -16(2)^2 + 112(2)

=-64+224

=160

max height enter in vertex formula: (-b\2a)

-(112)/2(-16)

=112/32

=3.5 (x-value) substitute in original equation and thats the maximum height.

s(3.5)= -16(3.5)^2+ 112(3.5)

=196

ball reaches gound:

need to find when s(t)=0; (use the quadratic formula or factor)

-16t^2 + 112t =0

-16t (t-7) =0

-16t =0 or t-7 =0

t =0 t =7