Question

You throw a rock straight upward at 30m/approximately how long does it rise? how long until it stops at the top of its flight?

Answer 1

Answer on Question #46140, Physics, Mechanics | Kinematics | Dynamics

You throw a rock straight upward at $30\mathrm{m/s}$. approximately how long does it rise? How long until it stops at the top of its flight?

Solution:

Given:

$$\begin{array}{l} v_0 = 30 \, \mathrm{m/s}, \\ v_f = 0, \\ t = ?, \\ h = ? \end{array}$$

The kinematic equation that describes an object's motion is:

$$2gh = v_f^2 - v_0^2$$

where $g = -9.81 \, \mathrm{m/s^2}$ is acceleration, $h$ is coordinate, $v_0$ is initial velocity and $v_f$ is final velocity.

Thus,

$$h = \frac{-v_0^2}{-2g} = \frac{30^2}{2 \cdot 9.81} = 45.9 \, \mathrm{m}$$

The other kinematic equation is:

$$g = \frac{v_f - v_0}{t}$$

Thus,

$$t = \frac{v_f - v_0}{g} = \frac{0 - 30}{-9.81} = 3.06 \, \mathrm{s} \approx 3.1 \, \mathrm{s}$$

Answer: $t = 3.1 \, \mathrm{s}$, $h = 45.9 \, \mathrm{m}$.

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