Regarding projectile motion.....?

Question

a body is projected at an angle of 45 dgrees with the horizontal with an initial velocity of 30 m/sec. In how many seconds will it reach back the ground, and determine the horizontal distane it will travel before it strikes the ground.

Answer 1

Initially, Vx=Vy=30 * cos(45) = 21.21

the midpoint of flight will be when velocity = 0 and the end of flight will be when it reaches 21.21 again in the opposite direction

Vy= 21.21 + a * t = 21.21 + (-9.8) * t

set that equal to -21.21 ant t = 4.329 second flight

Vx is constant since there is no acceleration in the x direction.

distance = 21.21 * 4.329 = 91.83 m