Answer to Question #134161 in Mechanics | Relativity for leydiana n marroquin

Question

During a baseball game, a batter hits a high
pop-up.
If the ball remains in the air for 4.72 s, how
high above the point where it hits the bat
does it rise? Assume when it hits the ground
it hits at exactly the level of the bat. The
acceleration of gravity is 9.8 m/s
2
.
Answer in units of m.

Answer 1

$\displaystyle y = v_{oy}t -\frac{gt^2}{2}$ - this is the equation that describes vertical component of motion of the ball.

We know that time for reaching the peak is 1/2 of total time in air.

$t = 4.72/2 =2.36 \; s$

Differentiating the first equation with respect to t, we have

$v_y = v_{y0} - gt$

When the ball reaches the highest point of its trajectory, its velocity becomes zero in this point (the highest point is turn-off point, velocity before $y_{max}$ was pointing up, after $y_{max}$ it was pointing down, in $y_{max}$ it is zero). Putting this into previous equation, we have

$0 = v_{y0} - 9.8 \cdot 2.36$

$v_{0y} = 23.128$ m/s.

Now we can calculate $y_{max}$ :

$y_{max} = 23.128 \cdot 2.36 - \frac{9.8 \cdot 2.36^2}{2}=27.29104 \approx 27.3$ m.

Answer: $27.3$ m.