The figure is attached below.

The initial speed of the projectile can be decomposed on to the horizontal `x` and vertical `y` axes depending on the initial firing angle.

`V_0x =V_0*cos(alpha)`

`V_0y =V_0*sin(alpha)`

On the horizontal (`x` axis) there in no acceleration but on the vertical axis (`y` axis) a...

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The figure is attached below.

The initial speed of the projectile can be decomposed on to the horizontal `x` and vertical `y` axes depending on the initial firing angle.

`V_0x =V_0*cos(alpha)`

`V_0y =V_0*sin(alpha)`

On the horizontal (`x` axis) there in no acceleration but on the vertical axis (`y` axis) a constant acceleration will be present (the acceleration due to gravity `g` directed downwards). At maximum height the vertical speed will be zero

`V_y(hmax) =0 m/s`

If we take the positive direction on the `y` axis upwards we can write

`V_y(hmax) =V_0y -g*t`

`0 =V_0*sin(alpha) -g*t`

`t = ((V_0*sin(alpha))/g) = (60*sin(60))/9.81 =5.30 s`

**Answer: the time for the projectile to reach its maximum height is 5.30 s**