Question

A cricketer throws a ball at an angle of
30° with the horizontal having velocity
30ms-1. The maximum height attained by
the ball is(Take g = 10 ms-2) ​

Answer 1

Answer:

the answer is 30 degree

Answer 2

Given:

A cricketer throws a ball at an angle of 30° with the horizontal having velocity 30m/s.

To find:

Maximum height attained by the ball.

Calculation:

This is an example of GROUND-GROUND PROJECTILE.

Let maximum height be h ; the general expression for maximum height of a projectile is given as follows:

$\boxed{ \sf{ \therefore \: h = \dfrac{ {u}^{2} { \sin}^{2} ( \theta) }{2g} }}$

Putting available values in SI units:

$= > \sf{ \: h = \dfrac{ {(30)}^{2} { \sin}^{2} ( {30}^{ \circ} ) }{2g} }$

$= > \sf{ \: h = \dfrac{ {(30)}^{2} { \sin}^{2} ( {30}^{ \circ} ) }{2 \times 10} }$

$= > \sf{ \: h = \dfrac{ 900 \times { \sin}^{2} ( {30}^{ \circ} ) }{2 \times 10} }$

$= > \sf{ \: h = \dfrac{ 900 \times { (\frac{1}{2} )}^{2} }{2 \times 10} }$

$= > \sf{ \: h = \dfrac{ 900 \times \frac{1}{4} }{2 \times 10} }$

$= > \sf{ \: h = \dfrac{ 90 \times \frac{1}{4} }{2} }$

$= > \sf{ \: h = \dfrac{ 90 }{8} }$

$= > \sf{ \: h = 11.25 \: m}$

So, maximum height attained is 11.25 m.