Question

A stone thrown from ground level returns to the same level 4 seconds after. With what speed was the stone thrown? Take g=10m/s2

Answer 1

Answer on Question 55018, Physics, Mechanics | Kinematics | Dynamics

Question:

A stone thrown from ground level returns to the same level 4 seconds after. With what speed was the stone thrown? Take $g = 10 \, m/s^2$.

Solution:

We can find the initial velocity of the stone from the kinematic equation:

$$v = v_0 + g t_{rise},$$

where, $v = 0 \, m/s$ is the final velocity of the stone when it reaches the maximum height and then became returns to the ground level, $g = 10 \, m/s^2$ is the acceleration of gravity and $t_{rise}$ is the time when the stone reaches the maximum height (it is obviously that $t_{rise} = t/2$, where $t = 4s$ is the total time that the stone spent in air).

Let's take the direction of the $y$-axis upward. Then, we can rewrite our kinematic equation:

$$v_0 - g t_{rise} = 0,$$ $$v_0 = g t_{rise} = g \frac{t}{2} = 10 \frac{m}{s^2} \cdot \frac{4s}{2} = 20 \frac{m}{s}.$$

Answer:

$$v_0 = 20 \frac{m}{s}.$$

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