Question

A ball is thrown with a speed u, at an angle with the horizontal. At the highest point of its motion, the
strength of gravity is somehow doubled. Taking this change into account, the total time of flight of the
projectile is​

Answer 1

Given:

A ball is thrown with a speed u, at an angle with the horizontal. At the highest point of its motion, the strength of gravity is somehow doubled.

To find:

Total time of the Projectile

Calculation:

Time taken for the projectile to reach the highest point be t1 :

$\therefore \: t1 = \dfrac{u \sin( \theta) }{g}$

Time taken for the projectile to reach the ground from the highest point be t2 :

$\therefore \: h = \dfrac{1}{2} (2g) {(t2)}^{2}$

$\: = > \dfrac{ {u}^{2} { \sin}^{2} ( \theta) }{2g} = \dfrac{1}{2} (2g) {(t2)}^{2}$

$\: = > \dfrac{ {u}^{2} { \sin}^{2} ( \theta) }{g} = 2g {(t2)}^{2}$

$\: = > {(t2)}^{2} = \dfrac{ {u}^{2} { \sin}^{2} ( \theta) }{2 {g}^{2} }$

$\: = > t2 = \sqrt{ \dfrac{ {u}^{2} { \sin}^{2} ( \theta) }{2 {g}^{2} } }$

$\: = > t2 = \dfrac{ u\sin ( \theta) }{ g \sqrt{2} }$

So , total time taken :

$t1 + t2 = \dfrac{u \sin( \theta) }{g} + \dfrac{u \sin( \theta) }{g \sqrt{2} }$

$= > t1 + t2 = \dfrac{u \sin( \theta) }{g} \bigg(1+ \dfrac{1}{ \sqrt{2} } \bigg)$

$= > t= \dfrac{u \sin( \theta) }{g} \bigg(1+ \dfrac{1}{ \sqrt{2} } \bigg)$

So, final answer is:

$\boxed{ \sf{ t= \dfrac{u \sin( \theta) }{g} \bigg(1+ \dfrac{1}{ \sqrt{2} } \bigg)}}$