Question

A crow is flying horizontally with a constant speed of $2.70\mathrm{m}/\mathrm{s}$ when it releases a clam from its beak as shown in Figure 4.38. The clam lands on the rocky beach $2.10\mathrm{s}$ later. Just before the clam lands, what is (a) its horizontal component of velocity and (b) its vertical component of velocity?

Short answer

Horizontal velocity: 2.70 m/s, Vertical velocity: 20.601 m/s.

Step-by-step solution

Understand the Problem

A crow is flying with a constant horizontal speed and releases a clam which follows projectile motion. We need to find the horizontal and vertical components of the clam's velocity just before it hits the ground.

Identify Initial Horizontal Velocity

Since the clam is released from the crow's beak, it inherits the horizontal velocity of the crow. Thus, the initial horizontal velocity, $v_{x0}$, is $2.70\mathrm{m}/\mathrm{s}$.

Constant Horizontal Velocity

In projectile motion, there is no horizontal acceleration (assuming air resistance is negligible), so the horizontal component of velocity remains constant throughout the motion. Thus, $v_x=2.70\mathrm{m}/\mathrm{s}$.

Calculate Time of Flight

The time of flight is given as $2.10\mathrm{s}$, which is the time the clam takes from being released to landing.

Determine Vertical Component of Velocity

The initial vertical velocity, $v_{y0}$, is $0\mathrm{m}/\mathrm{s}$ since the clam is not initially moving up or down. Using the equation $v_y=v_{y0}+gt$ where $g=9.81\mathrm{m}/{\mathrm{s}}^2$ (acceleration due to gravity) and $t=2.10\mathrm{s}$, we find the vertical velocity just before landing: $v_y=0+9.81\times 2.10=20.601\mathrm{m}/\mathrm{s}$.

Summarize Velocity Components

The horizontal velocity component just before landing is $2.70\mathrm{m}/\mathrm{s}$ and the vertical velocity component is $20.601\mathrm{m}/\mathrm{s}$.

Key concepts

Horizontal Velocity

When studying projectile motion, one of the important aspects is understanding the behavior of horizontal velocity. In the case of the crow releasing a clam while flying at a constant speed, the horizontal velocity is particularly straightforward. Since the crow is moving at a steady pace of $2.70\mathrm{m}/\mathrm{s}$ as it releases the clam, this velocity becomes the initial horizontal velocity of the clam.

In the absence of air resistance, which simplifies calculations in many physics problems, the horizontal velocity stays constant throughout the clam’s journey.

To break it down, here's why the horizontal velocity remains unchanged:

• There is no horizontal force acting on the clam after it leaves the crow's beak.
• In projectile motion, horizontal acceleration is zero.

Thus, the horizontal component of the clam's velocity just before it lands remains $2.70\mathrm{m}/\mathrm{s}$. Understanding this concept illustrates a key principle in physics: without an external force in the horizontal direction, velocity remains constant.

Vertical Velocity

Vertical velocity in projectile motion is more dynamic compared to horizontal velocity because it is directly influenced by gravity.

When the clam is initially released by the crow, it has no vertical velocity, or in other words, it isn't moving up or down yet, so the initial vertical velocity is $0\mathrm{m}/\mathrm{s}$.

However, once the clam is in free fall, gravity begins to accelerate it downward at $9.81\mathrm{m}/{\mathrm{s}}^2$.

To calculate the vertical velocity of the clam just before impact, we employ the equation for motion under constant acceleration:$$v_y=v_{y0}+gt$$where:

• $v_{y0}$ is the initial vertical velocity ($0\mathrm{m}/\mathrm{s}$ for the clam).
• $g$ is the acceleration due to gravity $9.81\mathrm{m}/{\mathrm{s}}^2$.
• $t$ is the time of flight, $2.10\mathrm{s}$.

Plugging in the values gives:$$v_y=0+(9.81\mathrm{m}/{\mathrm{s}}^2)\times (2.10\mathrm{s})=20.601\mathrm{m}/\mathrm{s}$$Thus, the clam’s vertical velocity just before it hits the beach is $20.601\mathrm{m}/\mathrm{s}$, moving downward due to the influence of gravity.

Acceleration Due to Gravity

Gravity is a key player in the story of projectile motion. It's this mysterious force that gives weight to objects and is responsible for the complex trajectory paths we see in projectiles like our clam.

Gravity exerts a constant acceleration on any object in free fall towards Earth, quantified as $9.81\mathrm{m}/{\mathrm{s}}^2$. This value tells us how much faster an object is moving vertically with each passing second.

In projectile motion, gravity specifically impacts the vertical component of velocity, causing even a horizontally launched object to eventually accelerate towards ground.

• This acceleration doesn't affect horizontal velocity; it only changes the vertical. That's why the horizontal velocity remains steady as we’ve discussed.
• Starting with no vertical motion, our clam is pulled down by gravity, gaining vertical speed every second it is airborne.

Understanding gravity's role is crucial because it allows us to predict how quickly an object, like the clam, will accelerate downward and how quickly it hits the ground. The immutable nature of gravitational acceleration ensures that once you know the time of flight, you can determine the vertical velocity just before landing.