Question

A ball is projected from ground such that its horixantal rangle is maximum

Answer 1

Answer:

sin(2*45) = U^2/g is the max range when theta = 45 deg is the elevation angle.

T = R/(Ucos(45)) is the time of flight for the first ball and t = sqrt(2R/g) for the second ball.

T/t = R/(Ucos(45))//sqrt(2R/g) = R/((Ucos(45))sqrt(2R/g)) = sqrt(R)/((Ucos(45))sqrt(2/g)) =U/sqrt(g)//((Ucos(45))sqrt(2/g)) = 1/(sqrt(g)cos(45)sqrt(2/g) = 1/(cos(45)sqrt(2)) = 1/(sqrt(2)/2 sqrt(2)) = 1//2/2 = 1. ANS...

Answer 2

sin(2*45) = U^2/g is the max range when theta = 45 deg is the elevation angle.

T = R/(Ucos(45)) is the time of flight for the first ball and t = sqrt(2R/g) for the second ball.

T/t = R/(Ucos(45))//sqrt(2R/g) = R/((Ucos(45))sqrt(2R/g)) = sqrt(R)/((Ucos(45))sqrt(2/g)) =U/sqrt(g)//((Ucos(45))sqrt(2/g)) = 1/(sqrt(g)cos(45)sqrt(2/g) = 1/(cos(45)sqrt(2)) = 1/(sqrt(2)/2 sqrt(2)) = 1//2/2 = 1. ANS...