A projectile is given an initial velocity of $$\left( {\hat i + 2\,\hat j} \right)\,m/s,$$ where $${\hat i}$$ is along the ground and $${\hat j}$$ is along the vertical. If $$g = 10\,m/{s^2},$$ the equation of its trajectory is:
A.
$$y = x - 5{x^2}$$
B.
$$y = 2x - 5{x^2}$$
C.
$$4y = 2x - 5{x^2}$$
D.
$$4y = 2x - 25{x^2}$$
Answer :
$$y = 2x - 5{x^2}$$
Solution :
$$\eqalign{
& \hat u = \hat i + 2\hat j = {u_x}\hat i + {u_y}\hat j \cr
& \Rightarrow u\,\cos \theta = 1,\,\,\,\,u\,\sin \theta = 2 \cr
& y = x\,\tan \theta - \frac{1}{2}\frac{{g{x^2}}}{{u_x^2}} \cr
& \therefore y = 2x - \frac{1}{2}g{x^2} = 2x - 5{x^2} \cr} $$
Releted MCQ Question on Basic Physics >> Kinematics
Releted Question 1
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