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
Answer:
the answer is step 2 I hope it's help
<span>6u^9+(8u^9+4v)
= </span><span>6u^9 + 8u^9 + 4v
= 14u^9 + 4v</span>
Answer:
no
Step-by-step explanation:
this is because it does not follow the pythogras theorem which states that a^2+b^2=c^2
