Question

for an object projected from ground with speed u horizontal range is two times the max. height attained.The horizontal range of object is

Answer 1

According to question,

Range (R) = 2 Height (H)
(u²sin2ß)/g = 2 x (u²sin²ß)/2g
2sinßcosß = sin²ß
tanß = 2/1 = P/B

With the help of Right angle triangle,

sinß=2/√5.......(1)
cosß = 1/√5.......(2)

Now, Horizontal range(R')= (u²sin2ß)/g
= (u²sinßcosß)/g
Now putting the value of sinß and cosß from (1) and (2) we get,
R' = 4u²/5g

Answer 2

Answer:

pls refer attachment

hope it helps!!!