A projectile is given an initial velocity of $(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}) \mathrm{ms}^{-1}$. The…
A projectile is given an initial velocity of $(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}) \mathrm{ms}^{-1}$. The equation of its path is $\left(g=10 \mathrm{~ms}^{-2}\right)$
$y=2 x-5 x^2$
$y=x-5 x^2$
$4 y=2 x-5 x^2$
$y=2 x-25 x^2$
Solution
Velocity of particle is $(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}) \mathrm{ms}^{-1}$ initially.
So, $\quad u_x=1 \mathrm{~ms}^{-1}$ and $u_y=2 \mathrm{~ms}^{-1}$
Also, $\quad a_x=0$ and $a_y=-10 \mathrm{~ms}^{-2}$
In time $t$,
Horizontal distance covered by projectile is
And vertical distance covered by projectile is
$
y=u_y t+\frac{1}{2} a_y t^2
$
Substituting the value of $t$ from Eq (i) in Eq (ii), we get
$
y=2 x-5 x^2
$
