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(C) In projectile motion,the horizontal displacement $x$ is given by $x = u_x t$,which is a linear function of time. The vertical displacement $y$ is

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Zero slope and negative curvature

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(C) In projectile motion,the horizontal displacement $x$ is given by $x = u_x t$,which is a linear function of time. The vertical displacement $y$ is given by $y = u_y t - \frac{1}{2} g t^2$,which is a downward-opening parabola. At the highest point,the vertical velocity $v_y = \frac{dy}{dt} = 0$. On a displacement-time graph for vertical motion,the slope is $\frac{dy}{dt}$. At the highest point,the slope is $0$. Since the path is a downward-opening parabola ($y = at^2 + bt + c$ with $a < 0$),the second derivative $\frac{d^2y}{dt^2} = -g$ is negative. However,the question asks for the displacement-time graph of the particle's motion. The vertical displacement $y(t)$ reaches a maximum at the peak,where the slope is $0$ and the curvature (concavity) is negative. Thus,the correct description is zero slope and negative curvature.

$A$ particle reaches its highest point when it has covered exactly one half of its horizontal range. The corresponding point on the displacement-time graph is characterized by

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