A particle is given an initial velocity of (i + 2ſ)ms, g = 10 ms?, th

1.	A particle is given an initial velocity of (i + 2ſ)ms, g = 10 ms?, the equation of trajectory is ay = x - 5x2b) y = 2x - 5x2 c) 4y = 2x - 5x?d) 4y = 2x - 25x2
Answer» Question :- A particle is given an initial velocity of $\sf \hat{i} + 2 \hat{j} \: \frac{m}{s} \: ,g = 10 \frac{m}{s}$ ,the equation of TRAJECTORY is - (A) y = x - 5 x² (B) y = 2x - 5x² (C) 4Y = 2x - 5x² (D) 4y = 2x - 25x² Answer :- $\implies \boxed{\sf \: y = 2x \: - 5 {x}^{2} } \:$ → OPTION (B) is correct . To find :- → Equation of trajectory Explanation :- Given that :- $\to \sf initial \: velocity \: (u) = (\sf \hat{i} + 2 \hat{j} )\: \frac{m}{s} \: \\ \\ \star \: \sf \: vertical \: component \: of \: initial \: velocity \:i s \\ \: \sf \: \: u_y = 2 \: \frac{m}{s} \: \\ \\ \star \sf \:horizontal\: component \: of \: initial \: velocity \:i s \\ \: \sf \: \: u_x = 1 \: \frac{m}{s} \: \:$ Hence,apply equation of motion for horizontal component ( x - axis ) $\implies \boxed{\sf \: s = ut + \frac{1}{2} a {t}^{2} } \\ \therefore \\ \\ \implies \sf \: x = u_x \: t - \frac{1}{2} a \: {t}^{2} \: \: \: \: \: \: \: \: \because \: \: a_x = 0 \: \\ \\ \implies \sf x = u_x \: t \: \: \: \: \: \: \: \: \because \: u_x = 1 \\ \\ \to \boxed{ \sf x = t}$ Now, apply SECOND equation of motion for vertical component - $\to \sf y = u_y \: t + \frac{1}{2} a_y \: {t}^{2} \\ \\ \because \sf u_y= 2 \: \: and \: a_y = - g = - 10 \\ \\ \to \sf y = 2t \: - \frac{1}{2} \times 10 \times {t}^{2} \\ \\ \to \sf \: y = 2t \: - 5 {t}^{2} \\ \\ \because \sf \: x = t \\ \\ \implies \boxed{\sf \: y = 2x \: - 5 {x}^{2} }$

A particle is given an initial velocity of (i + 2ſ)ms, g = 10 ms?, the equation of trajectory is ay = x - 5x2b) y = 2x - 5x2 c) 4y = 2x - 5x?d) 4y = 2x - 25x2

Answer»

Question :-

A particle is given an initial velocity of $\sf \hat{i} + 2 \hat{j} \: \frac{m}{s} \: ,g = 10 \frac{m}{s}$ ,the equation of TRAJECTORY is -

(A) y = x - 5 x²

(B) y = 2x - 5x²

(D) 4y = 2x - 25x²

Answer :-

$\implies \boxed{\sf \: y = 2x \: - 5 {x}^{2} } \:$

→ OPTION (B) is correct .

To find :-

→ Equation of trajectory

Explanation :-

Given that :-

$\to \sf initial \: velocity \: (u) = (\sf \hat{i} + 2 \hat{j} )\: \frac{m}{s} \: \\ \\ \star \: \sf \: vertical \: component \: of \: initial \: velocity \:i s \\ \: \sf \: \: u_y = 2 \: \frac{m}{s} \: \\ \\ \star \sf \:horizontal\: component \: of \: initial \: velocity \:i s \\ \: \sf \: \: u_x = 1 \: \frac{m}{s} \: \:$

Hence,apply equation of motion for horizontal component ( x - axis )

$\implies \boxed{\sf \: s = ut + \frac{1}{2} a {t}^{2} } \\ \therefore \\ \\ \implies \sf \: x = u_x \: t - \frac{1}{2} a \: {t}^{2} \: \: \: \: \: \: \: \: \because \: \: a_x = 0 \: \\ \\ \implies \sf x = u_x \: t \: \: \: \: \: \: \: \: \because \: u_x = 1 \\ \\ \to \boxed{ \sf x = t}$

Now, apply SECOND equation of motion for vertical component -

$\to \sf y = u_y \: t + \frac{1}{2} a_y \: {t}^{2} \\ \\ \because \sf u_y= 2 \: \: and \: a_y = - g = - 10 \\ \\ \to \sf y = 2t \: - \frac{1}{2} \times 10 \times {t}^{2} \\ \\ \to \sf \: y = 2t \: - 5 {t}^{2} \\ \\ \because \sf \: x = t \\ \\ \implies \boxed{\sf \: y = 2x \: - 5 {x}^{2} }$