WAEC Physics 2018 Paper

Concept:In projectile motion, the horizontal component of velocity remains constant because air friction is neglected and there is no horizontal acceleration.Explanation:The initial velocity is $15\,\mathrm{ms^{-1}}$ at an angle of $45^\circ$ to the horizontal.The horizontal component of the launch velocity is given by $U_x = U\cos\theta$.Substitute $U = 15\,\mathrm{ms^{-1}}$ and $\theta = 45^\circ$.So, $U_x = 15\cos 45^\circ = 15 \times 0.7071$.Hence, $U_x = 10.6065\,\mathrm{ms^{-1}} \approx 10.6\,\mathrm{ms^{-1}}$.At the highest point, the vertical velocity becomes $0\,\mathrm{ms^{-1}}$, but the horizontal velocity remains the same as the initial horizontal component because gravity acts only vertically.Answer:Magnitude of horizontal velocity at the highest point = $10.6\,\mathrm{ms^{-1}}$.Therefore, the correct option is C. $10.6\,\mathrm{ms^{-1}}$.