Question

The variation of horizontal displacement $x$ of a projectile with time $t$ is as shown in the figure above.
If the angle of projection is ${60}^o$, find the velocity of projection.

Answer 1

$V_x=u\cos \theta t$ where,

$\theta$-angle of projection

$u$-velocity of projection

$V_x$ in x-direction at any time

$\frac{dx}{dt}=\left(u\cos \theta t\right)$

Here, by graph $m=u\cos \theta =\frac{70}{4}$

$m=u\cos 60°=\frac{70}{4}$

$u=\frac{70}{2}=35m/s$