Question

Mcq a student is standing on a train travelling along a straight horizontal track at a speed of 10 m per second the student throws a ball into the air along a path that he sees to make an initial angle of 60 degree with the horizontal along the track the professor standing on the ground observes the ball to rise vertically the maximum height reached by ball is

Answer 1

Given:

Velocity of train on a straight horizontal track is 10 m/s. The makes an initial angle of 60° with the horizontal.

To find:

Max height reached by the ball.

Calculation:

This is question involving Ground to Ground Projectile.

Since , the train is moving at a speed of 10 m/s, the horizontal component of velocity of the projectile will also be 10 m/s.

Let velocity of projection be v ;

$v \cos( \theta) = 10$

$= > v \cos( {60}^{ \circ} ) = 10$

$= > v \times \dfrac{1}{2} = 10$

$= > \: v = 20 \: m {s}^{ - 1}$

So, max height will be:

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

$= > h = \dfrac{ {(20)}^{2} { \sin}^{2}( {60}^{ \circ} ) }{2g}$

$= > h = \dfrac{ 400 \times { ( \dfrac{ \sqrt{3} }{2}) }^{2}}{2g}$

$= > h = \dfrac{ 400 \times \dfrac{3}{2} }{2g}$

$= > h = \dfrac{ 200 \times 3 }{2g}$

$= > h = \dfrac{ 100 \times 3 }{g}$

$= > h = \dfrac{ 100 \times 3 }{10}$

$= > h = 10 \times 3$

$= > h = 30 \: m$

So, final answer is:

$\boxed{ \sf{ \red{ h = 30 \: m}}}$