Question

The trajectory of a projectile is given by `y=x tantheta-(1)/(2)(gx^(2))/(u^(2)cos^(2)theta)`. This equation can be used for calculating various phenomena such as finding the minimum velocity required to make a stone reach a certain point maximum range for a given projection velocity and the angle of projection required for maximum range. The range of a particle thrown from a tower is define as the distance the root of the tower and the point of landing.
In the previous problem, what should be the corresponding projection angle.
A. `4.5m`
B. 4m
C. 5m
D. `3.75m`

Answer 1

Correct Answer - D
`tan theta = (u^(2))/(Rg) = 2`
`H = R tan theta - (1)/(2) g. (R^(2))/(u^(2)) (1 + tan^(2) theta)`
`= (u^(2))/(g) - (1)/(2) g. (R^(2))/(u^(2)) - (1)/(2) (u^(2))/(g) = 10 - 1.25 - 5 = 3.75 m`