Question

A brass ball is shot vertically upward from the surface of an atmosphere-free planet with an
initial speed of 20.0 m/s. One second later, the ball has an instantaneous velocity in the upward
direction of 15.0 m/s. How high does the ball rise?

Answer 1

Answer on Question 71890, Physics, Other

Question:

A brass ball is shot vertically upward from the surface of an atmosphere-free planet with an initial speed of $20\ \mathrm{m/s}$. One second later, the ball has an instantaneous velocity in the upward direction of $15\ \mathrm{m/s}$. How high does the ball rise?

Solution:

Let’s first find the acceleration due to gravity on the planet from the kinematic equation:

$$v = v_0 + g t.$$

Then, we get:

$$g = \frac{v - v_0}{t} = \frac{15\ \frac{\mathrm{m}}{\mathrm{s}} - 20\ \frac{\mathrm{m}}{\mathrm{s}}}{1.0\ \mathrm{s}} = -5.0\ \frac{\mathrm{m}}{\mathrm{s}^2}.$$

The sign minus indicates that the acceleration due to gravity directed downward to the surface of the planet.

Then, we can find the time, $t_{\text{max}}$, that needs the ball to reach the maximum height from the same kinematic equation (also, at maximum height $v = 0$):

$$0 = v_0 + g t_{\text{max}},$$ $$t_{\text{max}} = -\frac{v_0}{-g} = -\frac{20\ \frac{\mathrm{m}}{\mathrm{s}}}{-5.0\ \frac{\mathrm{m}}{\mathrm{s}^2}} = 4.0\ \mathrm{s}.$$

Finally, we can find how high does the ball rise from another kinematic equation:

$$h_{\text{max}} = v_0 t_{\text{max}} + \frac{1}{2} g t_{\text{max}}^2 = 20\ \frac{\mathrm{m}}{\mathrm{s}} \cdot 4.0\ \mathrm{s} - \frac{1}{2} \cdot 5.0\ \frac{\mathrm{m}}{\mathrm{s}^2} \cdot (4.0\ \mathrm{s})^2 = 40.0\ \mathrm{m}.$$

Answer:

$$h_{\text{max}} = 40.0\ \mathrm{m}.$$

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