A projectile is fired with an initial speed of 50 m/s . It is observed that its horizontal range is exactly three times its maximum height. The maximum height attained by the projectile is (Take g = 10 m/s ^2 )

Step-by-step solution

Let the angle of projection be . The horizontal range is R = u^2 (2 ) g = u^2 (2 ) g . The maximum height is H = u^2 ^2 2g . Given that R = 3H , u^2 (2 ) g = 3 ( u^2 ^2 2g ) Cancelling u^2 g from both sides (since 0 ), 2 = 3 2 = 4 3 From = 4 3 , we get = 4 5 . Now, substituting u = 50 m/s , = 4 5 , and g = 10 m/s ^2 into the maximum height formula: H = (50)^2 ( 4 5 )^2 2(10) H = 2500 16 25 20 H = 1600 20 = 80 m . Answer: 80 m