Question

A softball is tossed into the air at an angle of 56.9 degrees with the vertical (that would be 33.1 degrees with the horizontal). The initial velocity is 19.5 m/s. What is the maximum height of the softball?

Answer 1

Answer on Question 63297, Physics, Mechanics, Relativity

Question:

A softball is tossed into the air at an angle of 56.9 degrees with the vertical (that would be 33.1 degrees with the horizontal). The initial velocity is $19.5\ m/s$. What is the maximum height of the softball?

Solution:

Let's take the upwards as the positive direction. Then, we can find the maximum height of the softball from the kinematic equation:

$$v_f^2 = v_i^2 + 2ah_{\max},$$

here, $v_f$ is the final velocity of the softball at the maximum height, $v_i$ is the initial velocity of the softball, $a = -g = -9.8\ m/s^2$ is the acceleration of gravity, $h_{\max}$ is the maximum height.

Let's write the kinematic equation in projection on the axis $y$:

$$v_{fy}^2 = v_{iy}^2 - 2gh_{\max}.$$

At the maximum height $v_{fy} = 0\ m/s$, so we get:

$$h_{\max} = \frac{-v_{iy}^2}{-2g} = \frac{v_{iy}^2}{2g}.$$

The projection of the initial velocity of the softball on the axis $y$ can be found as follows:

$$v_{iy} = v_i \cos \theta,$$

here, $\theta = 56.9{}^\circ$ is the angle with the vertical.

Substituting $v_{iy}$ into the equation for the maximum height we get:

$$h_{\max} = \frac{(v_i \cos \theta)^2}{2g}.$$

Finally, we can calculate the maximum height of the softball:

$$h _ {m a x} = \frac {(v _ {i} c o s \theta) ^ {2}}{2 g} = \frac {\left(1 9 . 5 \frac {m}{s} \cdot c o s 5 6 . 9 {}^ {\circ}\right) ^ {2}}{2 \cdot 9 . 8 \frac {m}{s ^ {2}}} = 5. 7 8 m.$$

**Answer:**

$$h _ {m a x} = 5. 7 8 m.$$

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