Question

Trajectory of a body in a projectile motion is given by y = x-x^2/80 and y are in meters. Find maximum height of the projectile.

Answer 1

Answer:

Equation of Trajectory:

$\footnotesize \longrightarrow \sf y = x \tan( \theta) - \dfrac{gx^2}{ {2u}^{2} \cos^{2} ( \theta) } \qquad...(i)$

Given:

$\footnotesize \longrightarrow \sf y = x - \dfrac{x^2}{80 } \qquad...(ii)$

Comparing equation (I) and (II) we have:

$\footnotesize \longrightarrow \sf x= x \tan( \theta)$

$\footnotesize \longrightarrow \sf \tan( \theta) = \dfrac{x}{x}$

$\footnotesize \longrightarrow \sf \tan( \theta) = 1$

$\footnotesize \longrightarrow \sf \theta = \tan^{ - 1} ( 1 )$

$\footnotesize \longrightarrow \underline{ \underline{ \sf \theta = {45}^{ \circ}}}$

Now simplify equation (I) :

$\footnotesize \longrightarrow \sf y = x \tan( \theta) - \dfrac{gx^2}{ {2u}^{2} \cos^{2} ( \theta) }$

$\footnotesize \longrightarrow \sf y = x \tan( \theta) - \bigg(\dfrac{gx^2}{ {2u}^{2} \cos^{2} ( \theta) } \times \dfrac{ \tan ( \theta) }{ \tan ( \theta) } \bigg)$

$\footnotesize \longrightarrow \sf y = x \tan( \theta) \bigg \{ 1 - \dfrac{xg}{2 {u}^{2} { \cos }^{2} (\theta) \tan( \theta) } \bigg \}$

$\footnotesize \longrightarrow \sf y = x \tan( \theta) \bigg \{ 1 - \dfrac{xg}{2 {u}^{2} { \sin }({ \theta} )} \bigg \}$

$\footnotesize \longrightarrow \sf y = x \tan( \theta) \bigg \{ 1 - \dfrac{x}{ R} \bigg \} \qquad...(iii)$

By taking equation (ii) we have :

$\footnotesize \longrightarrow \sf y = x - \dfrac{x^2}{80 }$

$\footnotesize \longrightarrow \sf y = x \bigg(1 - \dfrac{x}{80} \bigg) \qquad... (iv)$

Now, compare equation (iv) by equation (iii)

$\footnotesize \longrightarrow \underline{ \underline{ \sf R = 80 \:m}}$

By relationship between horizon range and maximum height :

$\footnotesize \longrightarrow \sf H = \dfrac{R \tan(\theta)}{4}$

$\footnotesize \longrightarrow \sf H = \dfrac{80 \tan( {45}^{ \circ}) }{4}$

$\footnotesize \longrightarrow \sf H = \dfrac{80 \times 1 }{4}$

$\footnotesize \longrightarrow \sf H = \dfrac{80 \times 1 }{4}$

$\footnotesize \longrightarrow\underline{\boxed{\red{ \bf H = 20\: m}}}$

Answer 2

• please check the attached file