Answer:
The maximum height of the ball is 256m
Step-by-step explanation:
Given the equation of a pathway modelled as pathway can be modeled by h = -16t² + 128t
At maximum height, the velocity of the ball is zero.
velocity = dh/dt
velocity = -32t + 128
Since v = 0 at maximum height
0 = -32t+128
32t = 128
t = 128/32
t = 4seconds
The maximum height can be gotten by substituting t = 4 into the modelled equation.
h = -16t² + 128t
h = -16(4)²+128(4)
h = -16(16)+512
h = -256+512
h = 256m
The maximum height of the ball is 256m.
Since the equation should be [tex]h = -16t^2 + 128t[/tex]
Also we know that at the maximum height, the velocity of the ball should be zero
So,
[tex]velocity = dh/dt[/tex]
velocity = -32t + 128
Now
0 = -32t+128
32t = 128
t = 4seconds
Now
[tex]h = -16t^2 + 128t\\\\h = -16(4)^2+128(4)[/tex]
h = -16(16)+512
h = -256+512
h = 256m
hence, The maximum height of the ball is 256m.
Learn more about height here: https://brainly.com/question/21551182
