Answer:
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
