Question

A ball rolls off the edge of a horizontal tabletop, \(4.23 \mathrm{ft}\) high. It strikes the floor at a point \(5.11 \mathrm{ft}\) horizontally away from the edge of the table. ( \(a\) ) For how long was the ball in the air? (b) What was its speed at the instant it left the table?

Short answer

The ball was in the air for approximately \( 0.518 \) seconds and its speed when it left the table was approximately \( 9.886 \mathrm{ft/s} \).

Step-by-step solution

Calculate the Time the Ball was in the Air

To solve this problem, use the equation for vertical motion under constant acceleration due to gravity, \( d = vt + 0.5at^2 \). Since the ball starts from rest, its vertical velocity component \( v = 0 \) at the beginning. The distance \( d \) fallen under gravity is equal to the height of the table, \( 4.23 \mathrm{ft} \). The acceleration \( a \) is due to gravity, and given that gravity's acceleration is \( -32.2 \mathrm{ft/s^2} \), the minus sign indicates that the ball is moving downwards. Setting these values in the equation and solving for \( t \) will give the time that the ball was in the air.

Calculation

Setting the values in the equation as described above, \( t = \sqrt{\frac{2 \cdot d}{a}} = \sqrt{\frac{2 \cdot 4.23}{32.2}} = 0.518 \mathrm{s} \). The ball was in the air for 0.518 seconds.

Calculate the Initial Speed

We know the horizontal displacement of the ball, \( 5.11 \mathrm{ft} \), and the time it was in the air, \( 0.518 \mathrm{s} \). As the horizontal motion of the ball is uniform, we can use the speed equation \( v = \frac{s}{t}\) to find the initial speed. Here, \( s \) represents the space traveled, and \( t \) is the time.

Calculation

Substituting these into the above equation, the initial speed \( v = \frac{5.11}{0.518} = 9.886 \mathrm{ft/s} \) is obtained. Therefore, the initial speed of the ball as it left the table was approximately \( 9.886 \mathrm{ft/s} \).

Key concepts

Vertical Motion

When an object moves under the influence of gravity alone, it is said to experience vertical motion. In our exercise, the ball falls from the table, which means it only has vertical motion after leaving the table's edge.
Here, the vertical motion is described using the equation of motion:

• Let's break it down: the distance the ball falls, \(d\), equals the height of the table.
• The vertical velocity \(v\) at the start is zero because it begins from rest.
• The acceleration \(a\) involved is due to gravity and is negative, meaning it acts downward.

The formula used is \(d = vt + 0.5at^2\). That's how we calculate the time \(t\) the ball takes to hit the floor.
By rearranging the formula for \(t\) when \(v = 0\) and substituting the values, we solve for the time using the equation \(t = \sqrt{\frac{2 \cdot d}{a}}\). This kind of motion helps us understand why gravity affects how long objects stay in the air.

Acceleration Due to Gravity

Gravity plays a crucial role in projectile motion, particularly in vertical motion. Known as the acceleration due to gravity, it is a constant value that defines how fast an object accelerates towards the Earth.
On Earth, this constant is about \(32.2 \mathrm{ft/s^2}\).
This acceleration doesn't depend on the object's mass or its initial speed. Whether it's a ball or a heavy rock, they accelerate downwards at the same rate if dropped from the same height.
Here's why it matters in calculations:

• The negative sign in calculations, \(-32.2 \mathrm{ft/s^2}\), indicates the direction of acceleration—downward.
• This constant value simplifies equations related to vertical displacement and can be plugged into equations to find time and speed.
• Gravity continues to accelerate the object until it hits the ground, and it's a critical element in predicting how long an object remains airborne.

Understanding this concept helps comprehend why actions like dropping and throwing objects give similar results regarding hitting the floor.

Horizontal Displacement

Horizontal displacement describes how far an object travels sideways during projectile motion. In our example, the ball rolls off the table and travels a certain distance horizontally before hitting the ground.
This horizontal motion is distinct because it remains at a constant speed once the ball leaves the table, assuming we neglect air resistance.

• We measure this by the total distance it covers, calculated using the time of flight and initial speed.
• The formula used is \(v = \frac{s}{t}\), where \(s\) is the displacement, and \(t\) is the time in the air.
• For the ball, the initial horizontal speed caused it to move horizontally while it fell, contributing to the distance covered before hitting the ground.

This simple concept of horizontal displacement supports interpreting how objects maintain horizontal motion, even when falling vertically.