Question

an astronaut on a planet with no atmosphere drops a feather into a 4.56m deep crater and records that the feather falls freely for 1.30s. what is magnitude of free - fall acceleration on the planet? m/s² with what speed does the feather strike the bottom of the crater? m/s

Explanation:

Step1: Use free - fall displacement formula

The displacement formula for free - fall is $y = v_0t+\frac{1}{2}at^{2}$. Since the feather is dropped ($v_0 = 0$), the formula simplifies to $y=\frac{1}{2}at^{2}$, where $y$ is the displacement, $a$ is the acceleration, and $t$ is the time. We need to solve for $a$. Rearranging the formula gives $a=\frac{2y}{t^{2}}$.

Step2: Substitute values

We know that $y = 4.56m$ and $t = 1.30s$. Substituting these values into the formula $a=\frac{2y}{t^{2}}$, we get $a=\frac{2\times4.56}{(1.30)^{2}}=\frac{9.12}{1.69}\approx5.4m/s^{2}$.

Step3: Use velocity - time formula

The velocity - time formula for free - fall is $v = v_0+at$. Since $v_0 = 0$, the formula simplifies to $v = at$.

Step4: Substitute values for velocity

We know that $a\approx5.4m/s^{2}$ and $t = 1.30s$. Substituting these values into the formula $v = at$, we get $v=5.4\times1.30 = 7.02m/s$.

Answer:

For the magnitude of free - fall acceleration: $5.4m/s^{2}$
For the speed of the feather when it strikes the bottom: $7.02m/s$