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An object is thrown vertically upward. Its height t seconds after launch is \(h(t)=20t-5t^2\) metres. At what times will the object be at a height of 15 metres?

Answer and explanation

Correct answer: 1 s and 3 s

For a height of 15 metres, set \(20t-5t^2=15\). Rearranging gives \(t^2-4t+3=0\), or \((t-1)(t-3)=0\). Thus, \(t=1\) s and \(t=3\) s. The first time occurs while the object is rising, and the second while it is falling. Exam tip: check both positive roots because a projectile can reach the same height twice.

Related tags

Quadratic EquationsWord ProblemsProjectile MotionHeight-Time RelationWord Problems And ApplicationsMathematicsClass 10 Mcq

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