Ken griffey jr hits a baseball so that it leaves the bat with an initial speed of 37 m/s at an angle of 53.1 degrees above horizontal. when the ball is hit, it is 1.07 m above the ground. how much total time is does the ball spend in the air before hitting the ground? what is the maximum height of the ball relative to the ground? how far does the ball travel?

h = Vy t - 1/2 g t^2 = -1.07 m  final height of ball relative to being hit

Vy = 37 * sin 53.1 = 29.6 m/s     initial vertical velocity

h = 29.6 t - 9.8 /2 t^2

4.9 t^2 -29.6 t - 1.07 = 0      

Solving the quadratic for t = 6.08 sec  for time of flight

Vy - g t = 0   where t is time to reach max height

t = 29.6 / 9.8 = 3.02 sec  to reach max height

S = Vx * t      where t is the distance traveled

S = V0 cos 53.1 * 6.08 = 37 * .60 * 6.08 = 135 m about 443 ft

i think it’s a. this is because if they measured the mass of something with a mass of 150g, then their final answer would have to be approximately 150g. though all of their answers are almost 150, rosa and tran have the closest answers (or the exact answer) to the accurate one.

d is the correct answer