A projectile is given an initial velocity of \[(\widehat{i}+2\widehat{j})\]m/s where I is along the ground and j is along the vertical. If g= 10 \[m/{{s}^{2}}\], the equation of its trajectory is?

Hint:we are given initial velocity in vector form, so we can find the magnitude of the velocity in horizontal and vertical direction from this. This is a projectile and we know in case of a projectile motion there is no horizontal acceleration and the acceleration in vertical direction is acceleration due to gravity.

Complete step by step answer:
Given, initial velocity, u= \[(\widehat{i}+2\widehat{j})\] m/s
So, for horizontal direction, \[{{u}_{x}}=1m/s\]and for vertical direction, \[{{u}_{y}}=2m/s\]
To find the horizontal distance covered, since there is no acceleration in horizontal direction,
$x={{u}_{x}}t \\
\Rightarrow x=1\times t \\
\Rightarrow x=t \\$
In vertical direction there is constant acceleration acting downwards,
$y={{u}_{y}}t-\frac{g{{t}^{2}}}{2} \\
\Rightarrow y=2\times t-\frac{10{{t}^{2}}}{2} \\
\therefore y=2t-5{{t}^{2}}$

Thus, x=t, putting this in the above equation we get,\[ y=2x-5{{x}^{2}}\].

Additional Information:
A trajectory or flight path is the path that an object with mass in motion follows through space as a function of time. In classical mechanics, the mass might be a projectile or a satellite. For example, it can be an orbit, the path of a planet, asteroid, or comet as it travels around a central mass. Acceleration due to gravity depends on various factors such as height, depth as well as rotation of the planet.

Note: When the body is projected whether, from the ground or a certain height In projectile problems, there is always no acceleration in the horizontal direction but There exists a constant acceleration in the vertical direction whose direction is downwards and this is constant. This is the acceleration due to gravity.