# A projectile is fired into the air from the top of a 200 meter cliff above a valley. Its initial velocity is 60 m/s at 60 degrees above the horizontal. How much times does it take for the projectile to reach its maximum height? 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

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