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Grade: 11 

                        
a cricket ball is thrown at a speed of 28m/s in a direction 30*( 30 degrees) above the horizontal .Calculate its maximum height.
4 years ago

Answers : (2)

Shaswata Biswas
132 Points 
							
Here, velocity of projection, u = 28 m/s. Angle of projection, \theta = 30০. 
We have, maximum height attained by a projectile, H = \frac{u^{2}sin \theta}{2g} = \frac{28^{2}sin 30^{o}}{2*9.8} = 20 m
So, the maximum height attained by the ball is 20 m.
THANKS
4 years ago
Shaswata Biswas
132 Points 
							
It sould be,
      H = \frac{u^{2}sin^{2} \theta}{2g} = \frac{28^{2}sin^{2}30^{o}}{2*9.8} = 10 m


So, the maximum height attained by the cricket ball will be 10 .
Sorry for mistake.
THANKS
4 years ago
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