QPage 48-49: Q6: HC Verma: Concepts of Physics - Volume 1 Chapter-3 Rest and Motion Kinematics Page 48-49: Q6 Solution: At which point on its path a projectile has the smallest speed...
Students often mistakenly assume the speed is zero at the peak because the vertical velocity component is zero. However, the horizontal component $v_x$ remains constant and non-zero (unless $\theta = 90^\circ$), so the minimum speed is $v_x$, not zero.
At which point on its path a projectile has the smallest speed...
Multiple Choice Options
- A
- B
- C
- D
Step-by-Step Verified Solution
The Short Answer
The projectile has the smallest speed at the highest point of its trajectory (the peak of its path).
Why?
To understand this, we need to break the projectile's velocity into two independent components:
Component | Behavior |
Horizontal velocity (vₓ) | Constant throughout the motion (assuming no air resistance). It never changes. |
Vertical velocity (vᵧ) | Decreases while going up (due to gravity acting downward), becomes zero at the peak, and then increases in the downward direction while coming down. |
The total speed (magnitude of velocity) at any instant is given by:
v = √(vₓ² + vᵧ²)
At the highest point:
- vᵧ = 0 (the projectile momentarily stops moving vertically).
- So the total speed becomes:
v = √(vₓ² + 0) = vₓ
At any other point (before or after the peak):
- vᵧ ≠ 0 (either positive going up, or negative coming down).
- So the total speed is:
v = √(vₓ² + vᵧ²) > vₓ (since vᵧ² is positive)
Therefore, the speed is minimum when vᵧ = 0, which happens only at the highest point.
Key Points
Aspect | Detail |
Smallest speed | At the highest point of the trajectory |
Speed at the peak | Equal to the horizontal component of the initial velocity = u cosθ (where θ is the angle of projection) |
Why? | Vertical velocity becomes zero; only horizontal velocity remains |
Is it zero? | No, unless the projectile is thrown straight up (θ = 90°), in which case vₓ = 0 and speed at the peak is zero. |
Special Case
- If the projectile is thrown vertically upward (θ = 90°):
- Horizontal component = 0
- At the highest point, speed = 0 (it momentarily comes to rest).
- If the projectile is thrown horizontally (θ = 0°):
- The highest point is the starting point itself.
- So the smallest speed is at the initial point (which is also the highest point for that motion).
Final Answer
The projectile has the smallest speed at the highest point of its trajectory. At that instant, its vertical velocity is zero, and its speed is equal to the constant horizontal component of velocity, i.e., v = u cosθ (where θ is the angle of projection).