Projectile in motion

Question

A missile is fired with a launch velocity of 15000 ft. per sec. at a direction 32.47 degrees. Find the height to which it rises the time of flight and the horizontal distance.

Answer 1

Vup=Vdown=Vv (vertical component that is )

Vv=15000 sin(32.47 ) also since

Vv= gt or4 t=Vv / g

h=0.5 g t^2

h= 0.5 g (v/g)^2

h=  (v)^2 /(2g) This how high it will rise

t= Vv/g (time to reach its highest point

t(total)= 2 t = 2Vv/g

finally the total distance traveled is

S= Vh t(total)=2 Vh Vv/g

S= 2V^2 cos(32.47 )sin(32.47 )/g simplified using double angel formula to

S= V^2 sin(2x32.47 )/g

Just plug the numbers in