at what angle shoud a body be projected with a velocity 24m/s just to pass over an obstacle 16m high at a horizontal distance of 32m?

pls tell me da method.......

i wud have replied u ..

but i know that seeing such an easy question ..a lot of ppl wud be posting answers ..

and u know wat ,..i am right

use



y = xtan@ - gx^2/2v^2cos@

16=32tan@ - g(32)^2/2v^2cos@

then use 1/cos^@ = sec^@ = 1+tan^@

u get a quadratic in tan@

solve for it

u get ur answer