Question

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 1

The maximum height is 162.67 m.

Step-by-step explanation:

Suppose the total height is h.

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

`u=32.6m/s`

And the remaining height will be,

`h'=h-(2)/(3)h\\ h'=(1)/(3)h`

So, according to question the initial speed is,

`u=32.6m/s`

Acceleration in the upward direction is negative,

`a=-9.8m/s^(2)`

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

Now according to third equation of motion.

`v^(2) =u^(2) -2as`

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

`0^(2) =32.6^(2) +2(-9.8)(h)/(3) \\h=(3* 1062.76)/(2* 9.8)\\h=162.67 m`

Therefore, the maximum height is 162.67 m.