Question

The minimum speed of the projectile in plane to plane projectile motion is zero and this is the instant when the projectile is at its highest point.

• A. True
• B. False

Answer 1

The correct option is B False

Projectile motion:

The diagrammatic representation of the projectile motion of a projectile is given below:

1. Here, $u$ is the velocity of the projectile, $u_x$ is the horizontal component and $u_y$ is the vertical component of the velocity of the projectile at an instant.
2. Here, $\theta$is the angle of projection and $A$ is the point where the projectile has the maximum height.
3. The only acceleration acting on the projectile is the gravity. Hence, the direction of acceleration will be downwards.

Maximum Height:

1. In projectile motion, the value of the vertical component of the velocity $u_y$ becomes zero at the highest point because acceleration was acting on it in the opposite direction till it reaches the maximum height.
2. The value of the horizontal component of the velocity, $u_x$remains constant throughout the projectile motion.
3. The velocity vector at any point on the trajectory is calculated using the formula $\mathit{u}\mathbf{=}\sqrt{{\mathbf{\left(}{\mathbf{u}}_{\mathbf{x}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{+}{\mathbf{\left(}{\mathbf{u}}_{\mathbf{y}}\mathbf{\right)}}^{\mathbf{2}}}$.
4. At the highest point, $u_y=0$ and $u_x$ remains constant which is why the velocity at the highest point will be $u=\sqrt{{\left({\mathrm{u}}_{\mathrm{x}}\right)}^2+{\left(o\right)}^2}={\mathrm{u}}_x$.
5. Thus, the correct statement is that the minimum speed of the projectile in a plane to plane projectile motion is equal to the horizontal component of the launch velocity and this is the instant when the projectile is at its highest point.

Hence, the given statement is false.