An object is thrown vertically upward and has a speed of 32.6 m/s when it reaches two thirds of its maximum height above the launch point. Determine its maximum height.

Answer :

Answer:

The maximum height is 162.67 m.

Explanation:

Suppose the total height is h.

And at the height of 2/3h the speed of an object is,

[tex]u=32.6m/s[/tex]

And the remaining height will be,

[tex]h'=h-\frac{2}{3}h\\ h'=\frac{1}{3}h[/tex]

So, according to question the initial speed is,

[tex]u=32.6m/s[/tex]

Acceleration in the upward direction is negative,

[tex]a=-9.8m/s^{2}[/tex]

And the final speed will be v m/s which is 0 m/s.

Now according to third equation of motion.

[tex]v^{2} =u^{2} -2as[/tex]

Here, v is the final velocity, u is the initial velocity, a is the acceleration, s is the displacement.

[tex]0^{2} =32.6^{2} +2(-9.8)\dfrac{h}{3} \\h=\dfrac{3\times 1062.76}{2\times 9.8}\\h=162.67 m[/tex]

Therefore, the maximum height is 162.67 m.