Question

plzzzz help me......​

Answer 1

Trajectory Equation - Projectile Motion

Question:-

• A body is projected at an angle of $60^{\circ}$ with horizontal. Its Range is 9 m. When x = 6 m, the value of y for the projectile would be:

Answer:-

We have the data:

• $\sf \theta = 60^{\circ} \\\\ \sf R = 9\ m\\\\ y =\ ?\ at\ x = 6\ m$

• We will use the Trajectory Equation to directly relate x and y, but first we must find the initial velocity $\sf u$. And we will do this by using the formula of Range.

$\sf\displaystyle R = \frac{u^2\, sin\, 2\theta}{g}\\\\\\\sf \implies u^2 = \frac{gR}{2\, sin\, \theta \, cos\, \theta} \\\\\\\sf \implies u^2 = \frac{10\times 9}{2\times sin\, 60^{\circ}\times cos\, 60^{\circ}} \\\\\\\sf \implies u^2 = \frac{10\times 9}{2\times\frac{\sqrt{3}}{2}\times \frac{1}{2}} \\\\\\\sf \implies u^2 = \frac{180}{\sqrt{3}}$

Now, we can use the Trajectory Equation and substitute $\sf u^2$

$\sf\displaystyle y = x\, tan\, \theta - \frac{gx^2}{2u^2\, cos^2\, \theta}\\\\\\\sf\implies y=6\, tan\, 60^{\circ}-\frac{10\times 6^2}{2\times \frac{180}{\sqrt{3}}\times cos^2\, 60^{\circ}}\\\\\\\sf \implies y=6\sqrt{3}-\frac{10\times 36\times \sqrt{3}}{2\times 180\times \frac{1}{4}} \\\\\\ \sf\implies y=6\sqrt{3}-4\sqrt{3}\\\\\\\sf\implies {\Large\boxed{\sf y=2\sqrt{3}\ m}} \ \textsf{Option (2)}$

Thus, The Answer is Option (2): $\sf 2\sqrt{3}\ m$.

$\rule{320}{1}$

Alternative Method:-

This is a slightly different method. Here, we will directly modify the trajectory equation to use the parameters we already know.

Consider the Trajectory Equation:-

$\sf\displaystyle y=x\, tan\, \theta-\frac{gx^2}{2u^2\, cos^2\,\theta}\\\\\\ \textsf{Let's modify the second term to bring R}\\\\\\ \sf\implies y = x\, tan\, \theta-\frac{gx^2}{2u^2\, cos^2\,\theta}\times\frac{sin\,\theta}{sin\,\theta}\\\\\\\sf\implies y = x\, tan\, \theta-\frac{g}{2u^2\, sin\,\theta\, cos\,\theta}\times\frac{x^2\ sin\,\theta}{cos\,\theta}\\\\\\\sf\implies y = x\, tan\, \theta-\frac{ x^2\, tan\,\theta}{R}\\\\\\ \implies \boxed{\sf y=x\, tan\,\theta\left(1-\frac{x}{R}\right)}$

• This is another form of the trajectory equation, which uses Range instead of the initial velocity.

• We can use this another form directly for our question.

$\sf\displaystyle y=x\, tan\,\theta\left(1-\frac{x}{R}\right) \\\\\\ \implies \sf y = 6\, tan\, 60^{\circ}\left(1-\frac{6}{9}\right)\\\\\\ \implies \sf y = 6\sqrt{3}\left(1-\frac{2}{3}\right)\\\\\\ \implies \sf y=6\sqrt{3}\times \frac{1}{3}\\\\\\ \implies {\Large\boxed{\sf y=2\sqrt{3}\ m}}\ \textsf{Option (2)}$

This method made our calculations a lot simpler. Anyway, The Answer is Option (2)

Answer 2

Answer:

Answer:-

We have the data:

$\sf \theta = 60^{\circ} \\\\ \sf R = 9\ m\\\\ y =\ ?\ at\ x = 6\ m$

We will use the Trajectory Equation to directly relate x and y, but first we must find the initial velocity $\sf u$. And we will do this by using the formula of Range.

$\sf\displaystyle R = \frac{u^2\, sin\, 2\theta}{g}\\\\\\\sf \implies u^2 = \frac{gR}{2\, sin\, \theta \, cos\, \theta} \\\\\\\sf \implies u^2 = \frac{10\times 9}{2\times sin\, 60^{\circ}\times cos\, 60^{\circ}} \\\\\\\sf \implies u^2 = \frac{10\times 9}{2\times\frac{\sqrt{3}}{2}\times \frac{1}{2}} \\\\\\\sf \implies u^2 = \frac{180}{\sqrt{3}}$

Now, we can use the Trajectory Equation and substitute $\sf u^2$

$\sf\displaystyle y = x\, tan\, \theta - \frac{gx^2}{2u^2\, cos^2\, \theta}\\\\\\\sf\implies y=6\, tan\, 60^{\circ}-\frac{10\times 6^2}{2\times \frac{180}{\sqrt{3}}\times cos^2\, 60^{\circ}}\\\\\\\sf \implies y=6\sqrt{3}-\frac{10\times 36\times \sqrt{3}}{2\times 180\times \frac{1}{4}} \\\\\\ \sf\implies y=6\sqrt{3}-4\sqrt{3}\\\\\\\sf\implies {\Large\boxed{\sf y=2\sqrt{3}\ m}} \ \textsf{Option (2)}$

Thus, The Answer is Option (2): $\sf 2\sqrt{3}\ m$.

$\rule{320}{1}$

Alternative Method:-

This is a slightly different method. Here, we will directly modify the trajectory equation to use the parameters we already know.

Consider the Trajectory Equation:-

$\sf\displaystyle y=x\, tan\, \theta-\frac{gx^2}{2u^2\, cos^2\,\theta}\\\\\\ \textsf{Let's modify the second term to bring R}\\\\\\ \sf\implies y = x\, tan\, \theta-\frac{gx^2}{2u^2\, cos^2\,\theta}\times\frac{sin\,\theta}{sin\,\theta}\\\\\\\sf\implies y = x\, tan\, \theta-\frac{g}{2u^2\, sin\,\theta\, cos\,\theta}\times\frac{x^2\ sin\,\theta}{cos\,\theta}\\\\\\\sf\implies y = x\, tan\, \theta-\frac{ x^2\, tan\,\theta}{R}\\\\\\ \implies \boxed{\sf y=x\, tan\,\theta\left(1-\frac{x}{R}\right)}$

This is another form of the trajectory equation, which uses Range instead of the initial velocity.

We can use this another form directly for our question.

$\sf\displaystyle y=x\, tan\,\theta\left(1-\frac{x}{R}\right) \\\\\\ \implies \sf y = 6\, tan\, 60^{\circ}\left(1-\frac{6}{9}\right)\\\\\\ \implies \sf y = 6\sqrt{3}\left(1-\frac{2}{3}\right)\\\\\\ \implies \sf y=6\sqrt{3}\times \frac{1}{3}\\\\\\ \implies {\Large\boxed{\sf y=2\sqrt{3}\ m}}\ \textsf{Option (2)}$

This method made our calculations a lot simpler. Anyway, The Answer is Option (2)