Maximum height of a projectile
Use the vertex of a quadratic height model h(t) = -5t^2 + vt + h0 to find a maximum height. Unlimited questions, five difficulty levels, and a full worked solution every time, included free with weekly lessons.
Try one
A water balloon is launched from the ground. Its height h, in metres, after t seconds is modeled by the equation shown. What is the maximum height it reaches?
h(t) = −5t2 + 20t
Answer: 20
See one solved, step by step
A beach ball is launched from the ground. Its height h, in metres, after t seconds is modeled by the equation shown. What is the maximum height it reaches?
h(t) = −5t2 + 20t
📘 Worked solution
1t = −b / (2a) = −20 / (2 · (−5)) = 2The vertex of a downward-opening parabola gives the maximum - find the time it occurs at first.
2h(2) = −5(2)2 + 20(2)Substitute that time back into the height equation.
3= −20 + 40 = 20Evaluate each term and add them up.
4Maximum height: 20 mThat's the greatest height the beach ball reaches before falling back down.
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