Question

Playing shortstop, you pick up a ground ball and throw it to second base. The ball is thrown horizontally, with a speed of \(22 \mathrm{~m} / \mathrm{s}\), directly toward point \(A\) as shown in Figure 4.37. When the ball reaches the second baseman \(0.45 \mathrm{~s}\) later, it is caught at point B. (a) How far were you from the second baseman? (b) What is the distance of vertical drop, from \(A\) to \(B\) ?

Short answer

(a) 9.9 m; (b) 0.99 m drop.

Step-by-step solution

Identify Given Information

We know the following information:- Horizontal velocity of the ball, \(v = 22 \text{ m/s}\)- Time for the ball to reach the second baseman, \(t = 0.45 \text{ s}\)

Calculate Horizontal Distance

The horizontal distance \(d\) can be calculated using the formula for constant speed: \[d = v \times t\]Substitute the known values:\(d = 22 \text{ m/s} \times 0.45 \text{ s} = 9.9 \text{ m}\)So, you were \(9.9\) meters away from the second baseman.

Determine Vertical Drop

To find the vertical drop, we need to use the formula for the distance fallen in free-fall, \(h = \frac{1}{2} g t^2\), where \(g\) is the acceleration due to gravity \(9.8 \text{ m/s}^2\).\[h = \frac{1}{2} \times 9.8 \text{ m/s}^2 \times (0.45 \text{ s})^2\]\(h \approx \frac{1}{2} \times 9.8 \times 0.2025 = 0.99 \text{ m}\)So, the ball vertically dropped approximately \(0.99\) meters from \(A\) to \(B\).

Key concepts

Horizontal Velocity

Horizontal velocity refers to the constant speed at which an object moves along a horizontal path. It is a fundamental factor in projectile motion, where an object is launched into the air and follows a curved trajectory.
In the case of the baseball, the horizontal velocity is given as \(22 \text{ m/s}\), meaning that the ball travels at a constant speed of \(22 \text{ m/s}\) in the horizontal direction. This speed does not change as it moves towards the second baseman.
To find out how far the baseball travels, we use the formula for constant speed:

• Distance \(d = v \times t\)

With \(v\) being the horizontal velocity and \(t\) the time taken. Substituting \(22 \text{ m/s}\) for the speed and \(0.45 \text{ s}\) for the time, the ball travels a distance of \(9.9 \text{ meters}\) horizontally. This calculation shows how horizontal velocity and time work together to determine distance.

Vertical Drop

The vertical drop in projectile motion refers to how much an object falls vertically due to gravity, while it moves horizontally. This drop is caused by the force of gravity acting on the object during its flight.
Even though the baseball is thrown horizontally, it simultaneously falls towards the ground. We calculate the vertical drop using the formula for free fall:

• Vertical drop \(h = \frac{1}{2} g t^2\)

Here, \(g\) represents the acceleration due to gravity, which is \(9.8 \text{ m/s}^2\). By substituting \(0.45 \text{ s}\) for the time, the equation yields\[ h = \frac{1}{2} \times 9.8 \times (0.45)^2 \approx 0.99 \text{ m} \]This tells us that while the baseball travels horizontally, it drops approximately \(0.99 \text{ meters}\) vertically, demonstrating the simultaneous horizontal and vertical components of projectile motion.

Acceleration Due to Gravity

Acceleration due to gravity, often denoted by \(g\), is a key component in understanding projectile motion. It represents the rate at which an object accelerates towards the Earth's surface due to gravitational force.
On Earth, this value is roughly \(9.8 \text{ m/s}^2\). This means that in the absence of other forces, an object's velocity increases by \(9.8 \text{ m/s}\) every second it is in free fall.

Impact on the Baseball

In the exercise, gravity is what causes the baseball to drop \(0.99 \text{ meters}\) as it moves horizontally. Although the ball is thrown horizontally, it cannot overcome the pull of gravity, resulting in a vertical descent.
Gravity acts independently of any horizontal forces, causing the vertical component of motion to appear separate from the horizontal motion. Therefore, while the horizontal velocity remains constant, the vertical drop is dictated solely by gravity's pull on the ball, highlighting the dual nature of projectile trajectories.