Assertion- Two balls are projected such that they have same range R and have maximum heights `H_1` &`H_2` respectively then `R = 4 sqrt(H_1 H_2)`
Reason- `H_1 = (u^2 sin^2(theta))/(2g)`, `H_2 = (u^2 cos^2(theta))/(2g)`

A projectil has the same range (R ) when the maximum heitht attained by it is either H_1 or H_1 . Find the relation between R, H_1 and H_2 .

A particle is projected from the ground with velocity u at angle theta with horizontal. The horizontal rnage, maximum hieght and time of flight are R,H and T respectively. They are given by, R = (u^(2)sin 2 theta)/(g),H = (u^(2)sin^(2)theta)/(2g) and T = (2usin theta)/(g) Now keeping u as fixed, theta is varied from 30^(@) to 60^(@) . Then,

If R and h represent the horizontal range and maximum height respectively of an obliquie projectile, the (R^(2))/(8h)+2h represents

STATEMENT-1: For maximum height H of a projectile when the macimum rnage is R the only valid relation is that H is 50% of R . STATEMENT-2: R = (u^(2)sin 2 theta)/(g) and H = (u^(2) sin^(2) theta)/(2g) .

Two particles are projected obliquely from ground with same speed such that their range 'R' are same but they attain different maximum heights h_(1) and h_(2) then relation between R, h_(1) and h_(2) is:

Two balls are thrown simultaneously at two different angles so that both have equal ranges. If H_(1) and H_(2) be the maximum heights attained in two cases. Then the summation H_(1)+H_(2) is equal to

A stone projected with a velocity u at an angle q with the horizontal reaches maximum heights H_(1) . When it is projected with velocity u at an angle (pi/2-theta) with the horizontal, it reaches maximum height H_(2) . The relations between the horizontal range R of the projectile, H_(1) and H_(2) , is

A particle has same range for two angle of projection with same speed 10ms^(-1) .The corresponding maximum height attained by particle is H_(1) and H_(2) respectively.Then, H_(1)+H_(2) is? (g=10m/s^(2))