Question

if time of Flight of projectile is 20 seconds for a maximum Range what should be its velocity projection and maximum height attained​

Answer 1

Answer:

u=141.4 , H=500

Explanation:

time of flight= 2usinteta/g

time of flight for maximum projectile is 20s

angle of max range=45°

time of flight=2usinteta/g

20= 2usin(45)/10

20=2u×1/√2/10

when u solve u will get velocity u,

u=141.42

then we have to find height,

height(H)=u^2sin^2teta/2g

H=141.42^2×sin^2(45°)/2(10)

H=19993×1/2/20

H=499.9 (which is approximately 500)