Projectile Motion Projectiles launched at an angle

Velocity in the x direction is constant Velocity in the y direction changes due to gravity

vxvx vyvy v 

v x is constant, v y is equal and opposite at equal times from the highest point

Velocity in the x direction  v x = v i (cos  ) vivi  x = v i (cos  )  t v x =  x/  t

 vivi Velocity in the y direction v y = v i (sin  ) v y =  y/  t  y = v i (sin  )  t+1/2g  t 2

yy xx  t  y = v i (sin  )  t +1/2g  t 2 

1. I throw a basketball at an angle of 58° above the horizontal with a velocity of m/s. What is the a) time it is in the air? b) distance it goes? and c) maximum height? a)  y = v i (sin  )  t + 1/2g(  t) 2 = 0  t = -2v i (sin  )/g = -2(24.36m/s)(sin58°)/-9.81m/s 2 = 4.21s b)  x = v i (cos  )  t = (24.36m/s)(cos58°)(4.21s) = 54.3m c)  y max = v i (sin  )(1/2)  t + 1/2g(1/2  t) 2 = (24.36m/s)(sin58°)(2.105s) + 1/2g(2.105) 2 = 21.7m -9.81m/s 2