# You throw a baseball directly upward at time t = 0 at an initial speed of 13.9 m/s. What is the maximum height the ball reaches above where it leaves your hand? At what times does the ball pass through half the maximum height? Ignore air resistance and take g = 9.80 m/s2.

We can use the equation of motion to solve for this problem.

Since the ball was thrown upwards, the gravity of Earth will affect its motion. The initial velocity, u, of ball is 13.9 m/s. At the maximum height, the velocity of the ball will be 0 m/s (else it...

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We can use the equation of motion to solve for this problem.

Since the ball was thrown upwards, the gravity of Earth will affect its motion. The initial velocity, u, of ball is 13.9 m/s. At the maximum height, the velocity of the ball will be 0 m/s (else it will continue moving upwards).

Using, `V^2 = U^2 + 2as`

where, u is initial velocity, v is final velocity, a is acceleration and s is the distance,

0^2 = 13.9^2 + 2 (-9.8)s

or, s = (13.9^2)/(2 x 9.8) = 9.86 m.

Thus, the ball will reach a maximum height of 9.86 m.

Half of this height is 4.93 m.

We can use, s = ut + 1/2 at^2

and solve for time. Here, s = 4.93 m, u = 13.9 m/s, a = -9.8 m/s^2

Solving for time, we get t = 0.42 s.

(when we solve the quadratic equation, we will get two values of time, one for upward motion and other for downward motion of ball).

Hope this helps.

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