Gravity and time - Kinematics?

Question

A baseball is hit nearly straight up into the air with a speed of 33 m/s.

How high does it go?

How long is it in the air?

Answer 1

Vmax = V(0) -gt, where v(0) is the initial velocity, and V max is the final velocity which the ball reaches upwards when it's velocity becomes equal to zero.

V 0 = g*t

33 = 9.8*t

t = 33/9.8 = 3.367 seconds

The maximum height the ball may reach is after 3.367 seconds.

Y = V(0)*t -0.5*g*(t)^2

Y = v(0)*t -0.5*9.8*(3.367)^2

Y = 33*3.367 - 0.5*9.8*3.367^2

Y = 55.55 meters

The ball stays in the air an interval =2*t=2*3.367=6.734 seconds

Answer 2

initial v = 33 m/s

acceleration = -g = - 9.8 m/s^2

How Long in the Air:

y = v*t + 1/2*a*t^2

y = 33t - 4.9t^2

Since initial y = final y = 0, then:

0 = 33t - 4.9t^2

Using the quadratic formula: t = 6.73 seconds

How High:

Since the ball is in the air 6.73 seconds, then it must hit its apex at 1/2t seconds, or t = 3.367 s.

y = 33 (3.367) - 4.9 (3.367^2)

y = 55.56 m