A particle is projected from a level ground making an angle $\mathrm{\theta }$ with the horizontal (x-direction). The vertical (y) component of its velocity changes with its x coordinate according to the graph shown in figure. Calculate $\mathrm{\theta }$ (in degree). Take g = 10 ms^-2.

In projectile motion x-component of the velocity will remain constant and at the highest position of projectile y-component of the velocity should be zero.

Hence, from graph, ${\mathrm{u}}_{\mathrm{y}}=10\mathrm{m}/\mathrm{s}$

${\mathrm{v}}_{\mathrm{y}}$ becomes zero when $\mathrm{x}=\frac{\text{ Range }}{2}$

$\therefore \text{ Range }=20\mathrm{m} \therefore \frac{2{\mathrm{u}}_{\mathrm{x}}{\mathrm{u}}_{\mathrm{y}}}{\mathrm{g}}=20$

$\therefore {\mathrm{u}}_{\mathrm{x}}=10\mathrm{m}/\mathrm{s}\therefore \tan \mathrm{\theta }=\frac{{\mathrm{u}}_{\mathrm{y}}}{{\mathrm{u}}_{\mathrm{x}}}=1\therefore \mathrm{\theta }={45}^{\circ }$