Respuesta :
Answer:
The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.
Explanation:
Given that,
Initial velocity u= 128 ft/sec
Equation of height
[tex]h = 128t-32t^2[/tex]....(I)
(a). We need to calculate the maximum height
Firstly we need to calculate the time
[tex]\dfrac{dh}{dt}=0[/tex]
From equation (I)
[tex]\dfrac{dh}{dt}=128-64t[/tex]
[tex]128-64t=0[/tex]
[tex]t=\dfrac{128}{64}[/tex]
[tex]t=2\ sec[/tex]
Now, for maximum height
Put the value of t in equation (I)
[tex]h =128\times2-32\times4[/tex]
[tex]h=128\ ft[/tex]
(b). The number of seconds it takes the object to hit the ground.
We know that, when the object reaches ground the height becomes zero
[tex]128t-32t^2=0[/tex]
[tex]t(128-32t)=0[/tex]
[tex]128=32t[/tex]
[tex]t=4\ sec[/tex]
Hence, The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.
