Mathematics

Mathematics

Problem

The height hh (in metres) of a ball thrown vertically upwards after tt seconds is given by h(t)=−4t2+16t+2h(t) = -4t^{2}+16t+2. a. (i) Calculate the times when the ball is on the ground (h=0)(h = 0). b. (ii) Find the maximum height reached by the ball and the time at which this occurs. c. (iii) State the yy-intercept of the graph and interpret its meaning in this context. d. (iv) Sketch the graph of hh against tt for the entire flight, clearly marking the intercepts and the highest point.
Intuitive explanation

This is a quadratic motion problem where you're modeling the ball's position over time. The key insight is recognizing that the coefficients tell a physical story: the −4t2-4t^2 term captures gravity's downward pull, the 16t16t term is the initial upward velocity, and the constant 22 is where the ball starts. To find when the ball reaches its peak height, look for the vertex of the parabola—that's where the upward motion stops and gravity takes over.

Intuitive explanation

This is a quadratic motion problem where you're modeling the ball's position over time. The key insight is recognizing that the coefficients tell a physical story: the −4t2-4t^2 term captures gravity's downward pull, the 16t16t term is the initial upward velocity, and the constant 22 is where the ball starts. To find when the ball reaches its peak height, look for the vertex of the parabola—that's where the upward motion stops and gravity takes over.

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