Physics - Kinematics Question with Solution | TestHub

$y=10x-2x^2$

$$\frac{dy}{dt}=10\frac{dx}{dt}-4x\frac{dx}{dt}$$

$$v_y=10v_x-4x{v}_x$$

At the highest point, $v_y=0$. So, $10-4x=0$.

$$x=2.5$$

The range is $2.5\times 2=5m$. At $x=2.5m$, the particle is at its maximum height.

$$y=10\times \frac{5}{2}-2\times \frac{25}{4}$$

$$y=\frac{50}{2}-\frac{50}{4}$$

$$y=25-12.5$$

$$y=12.5m$$

At $x=2.25$:

$$v_y=10v_x-4\times 2.25v_x$$

$$v_y=10v_x-9v_x$$

$$v_y=v_x$$

$$\frac{v_y}{v_x}=1=\tan 45^{\circ }$$