# What is the distance a ball travels (measured along the ground) if its highest point on its trajectory is 54m from the ground and is inclined 20 degrees upwards from the throwing position?

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### 1 Answer

Assuming the path of the ball is a perfect parabola (this isn't necessarily true, since drag due to air resistance may change this, but this is the simple model of Galileo), then the distance along the ground as the ball ascends to its highest point is the base of a triangle with height 54m and inclination 20 degrees (note this is the angle from the throwing position to the highest point the ball reaches, and *not *the initial throwing angle which is `theta` on the quoted wikipedia page about 'trajectories'). The distance along the ground as the ball descends back down is the same distance again, since we're assuming its path is a perfect parabola.

Using trigonometry, the base of the triangle in metres is given by

`x = 54/(tan^(20))`

``since `tan(theta) = "O"/A` ``where `theta` is the angle of inclination, O (opposite side) is the height and A (adjacent side) is the length of the base (=`x`

We have then that x = A = 54/tan(20) = 54/0.364 = 148.36. Therefore the total distance along the ground travelled by the ball is

2x = 2A = **296.73m to 2dp.**

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