Question

The equation of the trajectory is given by (A) 180y = 240 x – x2 (B) 180 y = x2 – 240x (C) 180y = 135x – x2 (D) 180y = x2 – 135x

Answer 1

Correct Question:

A projectile is thrown with a velocity of 50 m/s at an angle of 53° with the horizontal. Find the equation of the trajectory.

Calculation:

Distance travelled in time t along X direction:

$\therefore \: x = u \cos( \theta) \times t$

$= > \: x = 50\cos( {53}^{ \circ} ) \times t \: \: \: \: .....(1)$

Distance travelled in time t along Y axis:

$\therefore \: y = u \sin( \theta) t - \dfrac{1}{2} g {t}^{2}$

$= > \: y = 50\sin( {53}^{ \circ} ) t - \dfrac{1}{2} g {t}^{2} \: \: \: \: ....(2)$

Putting value of "t" from eq.(1) into eq.(2):

$= > \: y = 50\sin( {53}^{ \circ} ) \times \bigg \{ \dfrac{x}{50 \cos( {53}^{ \circ} ) } \bigg \} - \dfrac{1}{2} g { \bigg \{\dfrac{x}{50 \cos( {53}^{ \circ} ) } \bigg \} }^{2}$

$= > \: y =x \tan( {53}^{ \circ} ) - \dfrac{g {x}^{2} }{2 {(50)}^{2} { \cos}^{2}( {53}^{ \circ} ) }$

$= > \: y = \dfrac{4x}{3} - \dfrac{g {x}^{2} }{2 {(50)}^{2} {( \frac{3}{5}) }^{2} }$

$= > \: y = \dfrac{4x}{3} - \dfrac{10 \times 25{x}^{2} }{2 \times 2500 \times 9}$

$= > \: y = \dfrac{4x}{3} - \dfrac{10 \times {x}^{2} }{2 \times 100 \times 9}$

$= > \: y = \dfrac{4x}{3} - \dfrac{{x}^{2} }{2 \times 10 \times 9}$

$= > \: y = \dfrac{4x}{3} - \dfrac{{x}^{2} }{180}$

$= > \: 180y = 240x - {x}^{2}$

So, the required Equation is:

$\boxed{ \bold{\: 180y = 240x - {x}^{2} }}$

Answer 2

Correct Question:-

A projectile is thrown with a velocity of 50 m/s at an angle of 53° with the horizontal. Find the equation of the trajectory.

We can use the direct formula for equation of trajectory since we have been given the initial velocity (u) and angle of projection $\sf (\theta).$

The equation of trajectory is:

${\green{\bigstar}} \ \ \boxed{\sf{y = x\:tan\theta - \dfrac{gx^2}{2u^2cos^2\theta}}}$

We must remember that,

$\boxed{\begin{minipage}{4.2cm} \quad \quad \bullet \sf{ sin\:53 = 4/5} \\ \\ \sf{\quad \quad \bullet cos\:53 = 3/5} \\ \\ \sf{\quad \quad \bullet tan\:53 = 4/3} \\ \\ \sf{(All\:\: angles\:\: are\:\: in\:\: degrees)} \end{minipage}}$

Hence, using the above formula, we can write the equation of trajectory as:

$\sf y = x\:tan\theta - \dfrac{gx^2}{2u^2cos^2\theta}$

$\longrightarrow \sf y = x\:tan\:53^o - \dfrac{10x^2}{2(50)^2\bigg(\dfrac{3}{5}\bigg)^2}$

$\longrightarrow \sf y = \dfrac{4}{3}x - \dfrac{1 {\cancel{0}} x^2}{ 180 {\cancel{0}} }$

$\longrightarrow \sf y = \dfrac{4}{3}x - \dfrac{x^2}{180}$

$\longrightarrow \sf y = \dfrac{240x - x^2}{180}$

$\longrightarrow \underline{\underline{\sf{\green{ 180y = 240x - x^2}}}}$