Question

A body of mass m, projected vertically upwards with an initial velocity 'u' reaches a maximum
height 'h'. Another body of mass m, is projected along an inclined plane making an angle 30' with
the horizontal and with speed 'u'. The maximum distance travelled along the incline is​

Answer 1

Given:

A body of mass m, projected vertically upwards with an initial velocity 'u' reaches a maximum

height 'h'. Another body of mass m, is projected along an inclined plane making an angle 30' with

the horizontal and with speed 'u'.

To find:

Max distance travelled along the inclined plane?

Calculation:

When thrown vertically:

$\therefore \: {v}^{2} = {u}^{2} + 2ah$

$= > \: {v}^{2} = {u}^{2} + 2( - g)h$

$= > \: {v}^{2} = {u}^{2} - 2gh$

$= > \: {(0)}^{2} = {u}^{2} - 2gh$

$= > \: 0 = {u}^{2} - 2gh$

$= > \: 2gh = {u}^{2}$

$\boxed{ = > \: h = \dfrac{ {u}^{2} }{2g} }$

When Projected along the incline:

• The acceleration will be $g\sin(\theta)$

• Let the distance travelled be s.

$\therefore \: {v}^{2} = {u}^{2} + 2as$

$= > \: {(0)}^{2} = {u}^{2} + 2 \{ - g \sin( \theta) \} s$

$= > \: {(0)}^{2} = {u}^{2} - 2 g \sin( \theta) s$

$= > \: 0 = {u}^{2} - 2 g \sin( \theta) s$

$= > \: 2 g \sin( \theta) s = {u}^{2}$

$= > \: s = \dfrac{ {u}^{2} }{2g \sin( \theta) }$

$= > \: s = \dfrac{ {u}^{2} }{2g } \times \dfrac{1}{ \sin( \theta) }$

$= > \: s = h \times \dfrac{1}{ \sin( \theta) }$

$= > \: s = \dfrac{h}{ \sin( \theta) }$

$= > \: s = \dfrac{h}{ \sin({30}^{\circ}) }$

$= > \: s = \dfrac{h}{\frac{1}{2}}$

$=>\: s = \dfrac{2h}{1}$

$=>\: s = 2h$

So, final answer is:

$\boxed{\bold{\: s =2h }}$