Question

Find the angle of projectile for a body to have same horizontal range and maximum height .....

Answer 1

I Hope it will help u

Answer 2

hello mate here is ur answer

let tita , be the angle of projection for which both horizontal range and maximum height of projectile are equal , now

since,

H=R===(1)


, H =u^2sin^2tita / 2g=====(2)

R= u^2sin2tita/g======(3)


from equation 1 we are having
H=R

therefore

substituting equation 2 and 3
we get

u^2sin^2tita/2g= u^2sin2tita/g


after cancelling out g and u^2

left out term will be

sin^2tita/2= sin2tita

sin^2tita = 2(2sintita×costita)

sin^2 tita = 4×sintita×costita

sintita=4costita


this implies


sintita/costita= 4 ====(4)

since sin tita / cos tita = tan tita

equation four implies

tantita=4

therfore, tita = tan^-1(4)


here's the answer,.

hope it helps u


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