P0295 Mechanics › Kinematics

Projectile Minimum Speed vs Flight Time

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Kinematics Intermediate projectile

Source: Principles of Physics

Problem

A projectile is launched from level ground with a constant initial speed $v_0$. The launch angle $\theta_0$ can be varied. The maximum possible range is $R_{max}$, and the maximum possible flight time is $t_{max}$.

Determine the minimum speed of the projectile during its flight if the launch angle is chosen such that the flight time is $t = \alpha t_{max}$, where $\alpha$ is a constant factor ($0 < \alpha \le 1$). Express your answer in terms of $R_{max}$, $g$, and $\alpha$.
Problem image

[Q1] $v_{min} = \sqrt{gR_{max}(1 - \alpha^2)}$

The range of a projectile is $R = (v_0^2 \sin(2\theta_0))/g$. The maximum range $R_{max}$ occurs when $\sin(2\theta_0) = 1$ (at $\theta_0 = 45^\circ$), giving $R_{max} = v_0^2/g$. The initial speed is therefore $v_0 = \sqrt{gR_{max}}$.

The total flight time is $t = (2v_0 \sin\theta_0)/g$. The maximum flight time $t_{max}$ occurs when $\sin\theta_0 = 1$ (at $\theta_0 = 90^\circ$), so $t_{max} = 2v_0/g$.

The problem states that the flight time is a fraction $\alpha$ of the maximum:

$$ t = \alpha t_{max} \implies \frac{2v_0 \sin\theta_0}{g} = \alpha \left(\frac{2v_0}{g}\right) $$

This simplifies to $\sin\theta_0 = \alpha$.

The minimum speed during flight, $v_{min}$, occurs at the apex of the trajectory where the vertical velocity is zero. This speed is equal to the constant horizontal component of velocity, $v_x$.

$$ v_{min} = v_x = v_0 \cos\theta_0 $$

Using the identity $\cos\theta_0 = \sqrt{1-\sin^2\theta_0}$, we find $\cos\theta_0 = \sqrt{1-\alpha^2}$.

Substituting the expressions for $v_0$ and $\cos\theta_0$ into the equation for $v_{min}$:

$$ v_{min} = (\sqrt{gR_{max}})(\sqrt{1-\alpha^2}) = \sqrt{gR_{max}(1-\alpha^2)} $$