Question

A pilot performs an evasive maneuver by diving vertically at 310 m/s. If he can withstand an acceleration of 9.0g's without blacking out, at what altitude must he begin to pull out of the dive to avoid crashing into the sea

Answer 1

Answer on Question #48870 – Physics – Mechanics | Kinematics | Dynamics

1. A pilot performs an evasive maneuver by diving vertically at $310~\mathrm{m/s}$. If he can withstand an acceleration of $9.0\mathrm{g/s}$ without blacking out, at what altitude must he begin to pull out of the dive to avoid crashing into the sea?

Assuming the fly of the airline to be the motion with constant acceleration, we can connect the displacement and velocities and the acceleration: $s = \frac{v_1^2 - v_0^2}{2a}$, where $v_0$ and $v_1$ are the initial and final velocity.

The necessary altitude is $h = -s$. So, $\boxed{h = \frac{v_0^2 - v_1^2}{18g}}$.

Let check the dimension: $\left[h\right] = \frac{(m / s)^2}{m / s^2} = m$.

Let evaluate the quantity: $h = \frac{310^2 - 0^2}{18 \cdot 9.81} = 544(m)$.

Answer: $544m$.

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