Question

. A particle is projected with velocity u at an angle θ with the vertical. Show that its path is parabolic. If the velocity of the projection of the particle is doubled, what is the percentage change in maximum height attained by projectile​

Answer 1

Particle projected at angle $\theta$ with velocity u.

Now, we can say :

Displacement along X axis:

$\sf x = u \cos( \theta) \times t$

Displacement along Y axis:

$\sf y = u \sin( \theta)t - \dfrac{1}{2}a {t}^{2}$

Now, putting value of t in eq.(2):

$\sf \implies y = u \sin( \theta)t - \dfrac{1}{2}a {t}^{2}$

$\sf \implies y = \bigg \{u \sin( \theta) \times \dfrac{x}{ \cos( \theta) } \bigg \} - \dfrac{1}{2}a { \bigg \{ \dfrac{x}{ \cos( \theta) } \bigg \}}^{2}$

$\boxed{ \sf \implies y = x \tan( \theta) - \dfrac{g {x}^{2} }{2 {u}^{2} { \cos}^{2}( \theta) } }$

HENCE, TRAJECTORY IS PARABOLIC !

Now, max height formula is :

$\rm h = \dfrac{ {u}^{2} { \sin}^{2} ( \theta)}{2g}$

Now, for double velocity:

$\rm h_{2} = \dfrac{ {(2u)}^{2} { \sin}^{2} ( \theta)}{2g}$

$\rm \implies h_{2} =4 \times \dfrac{ {u}^{2} { \sin}^{2} ( \theta)}{2g}$

$\rm \implies h_{2} =4 h$

So, % change :

$\rm\% \: change = \frac{4h - h}{h} \times 100$

$\rm \implies\% \: change = 300 \%$

Hope It Helps.