Answer:
[tex]v(t) = 256 - 32t[/tex]
[tex]y_{\rm max} = 1024 ~{\rm m}[/tex]
Explanation:
The equation of kinematics will be used to find the velocity function.
[tex]v = v_0 + at\\v(t) = v_0 + gt[/tex]
where g is equal to -32 ft/sec2 and v0 is equal to 256 ft/sec.
Therefore, the velocity function is
[tex]v(t) = 256 - 32t[/tex]
The position function is the antiderivative of the velocity function.
[tex]y(t) = \int{v(t)} dt = 256t - 16t^2[/tex]
The greatest height is where the derivative of the position function is equal to zero. So,
[tex]\frac{dy(t)}{dt} = v(t) = 0\\0 = 256 - 32t\\t = 8~{\rm sec}[/tex]
Putting this into the position function gives the greatest height.
[tex]y(t=8) = 256(8) - 16(8^2) = 1024~m[/tex]
The greatest height attained by the object is 1024 ft.
We have to use the equations of motion under gravity;
v(t) = u - gt
The negative shows that the motion occurred in a direction that opposes gravity.
When;
u = 256 ft/sec
g = 32 ft/sec2
t = ?
v = 0 ft/sec (at maximum height)
t = u/g = 256 ft/sec/32 ft/sec2
t = 8 s
Now, we can obtain the greatest height by antiderivative as follows;
v(t) = u - gt
v(t) = 256 - 32t
Integrating with respect to t;
∫v(t) dt= ∫256 - 32t dt
h = ∫256 - 32t dt
h = 256t - 32t^2/2
h = 256t - 16t^2
Where t = 8
h = (256 × 8) - (16 × 8^2)
h = 2048 - 1024
h = 1024 ft
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