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A projectile is launched directly upwards and its height, H metres above the ground is given by H=20t-5t² where t is the time in seconds. After how many seconds will the projectile be 20 metres about the ground?

When the height is 20, that means H = 20.

This gives the equation 20 = 20t - 5t².

Adding 5t² - 20t to both sides gives 5t² - 20t + 20 = 0.

Dividing by 5 gives t² - 4t + 4 = 0.

Factoring gives (t-2)(t-2) = 0.

The solution is where t-2 = 0, or at t=2.

Putting that back in gives 20(2) - 5(2²) = 40 - 5*4 - 40 - 20 = 20,

so that's the right answer.

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