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In this video, we calculate the horizontal range and maximum height of a kicked ball.

Question one. A ball is kicked at twenty metres per second at forty-five degrees to the ground. How far does it travel before it lands, and how high does it go? G equals nine point eight metres per second squared.

Assume the ball lands at its launch height and air resistance is negligible. The initial speed is twenty metres per second, and the launch angle is forty-five degrees. The downward acceleration is nine point eight metres per second squared.

You need the horizontal range and the maximum height above the launch point.

First, resolve the initial velocity into horizontal and vertical components using cosine and sine. The horizontal component is twenty times cosine forty-five degrees, giving ten root two metres per second. The vertical component is twenty times sine forty-five degrees, also giving ten root two metres per second.

Because launch and landing heights match, the total flight time is twice the initial vertical component divided by g.

Substituting twice ten root two metres per second over nine point eight metres per second squared gives two point eight eight six seconds.

Horizontal velocity stays constant, so the range is horizontal velocity multiplied by flight time.

Multiplying ten root two metres per second by two point eight eight six seconds gives forty point eight metres. At maximum height, vertical velocity is zero, so the height is the initial vertical component squared divided by twice g.

Substitute ten root two metres per second squared, divided by nineteen point six metres per second squared.

The maximum height is ten point two metres. As a check, at forty-five degrees the range should be four times the maximum height, which these values satisfy.

Hence, the ball travels forty point eight metres horizontally and rises ten point two metres above its launch point.