Question

[T] You throw a ball of mass 1 kilogram upward with a velocity of \(a=25 \mathrm{~m} / \mathrm{s}\) on Mars, where the force of gravity is \(g=-3.711 \mathrm{~m} / \mathrm{s}^{2}\). Use your calculator to approximate how much longer the ball is in the air on Mars.

Short answer

The ball is in the air approximately 13.46 seconds on Mars.

Step-by-step solution

Determine Time to Reach Maximum Height

When the ball reaches its maximum height, its velocity is 0 m/s. We use the equation of motion \( v = u + at \), where \( v \) is the final velocity, \( u \) is the initial velocity (25 m/s), \( a \) is the acceleration due to gravity (-3.711 m/s\(^2\)), and \( t \) is the time. Set \( v = 0 \): \[ 0 = 25 - 3.711t \]Solve for \( t \) to find the time to reach maximum height:\[ t = \frac{25}{3.711} \approx 6.73 \text{ seconds} \]

Calculate Total Time in the Air

The total time the ball is in the air is twice the time to reach maximum height, because the time taken to ascend is equal to the time taken to descend. Thus, the total time \( T \) is:\[ T = 2t = 2 \times 6.73 \approx 13.46 \text{ seconds} \]

Key concepts

Equations of Motion

Equations of motion are essential tools when analyzing projectile motion. They help us understand how objects move under the influence of forces like gravity.
The primary equations include:

• The formula for velocity: \( v = u + at \). Here, \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time taken. It's useful when determining the time it takes for an object to stop moving upwards.
• The displacement equation: \( s = ut + \frac{1}{2}at^2 \). Although not directly used in our solution, it's important for determining distances traveled.
• The equation for final velocity in terms of displacement: \( v^2 = u^2 + 2as \). It helps find final velocity when displacement is known.

Using these equations correctly allows us to solve problems, such as calculating the time a ball is in the air or finding its maximum height.

Gravity on Mars

Gravity differs on various planets, affecting motion significantly. On Earth, gravity is roughly \( 9.81 \, \text{m/s}^2 \). However, Mars has much weaker gravity, about \( 3.711 \, \text{m/s}^2 \).
This reduced gravitational pull means that objects thrown on Mars will ascend higher and be airborne longer compared to Earth.
Understanding gravitational differences is crucial for space missions and experiments. It helps adjust expectations for movement and forces needed for various tasks on other planets.
On Mars:

• Low gravity means less force needed to lift objects.
• Projectiles travel higher and stay longer in the air.
• Unique challenges and advantages in designing equipment.

Velocity and Acceleration

Velocity and acceleration are key concepts for understanding projectile motion. Velocity refers to how fast something moves in a particular direction, while acceleration shows how quickly velocity changes.
In the case of the ball thrown on Mars:

• It starts with an initial velocity of \( 25 \, \text{m/s} \).
• The acceleration impacts velocity, here due to Mars' gravity, at \( -3.711 \, \text{m/s}^2 \).
• Negative acceleration slows the ball until its velocity becomes zero at the peak height.

This scenario highlights how acceleration due to gravity continually decreases upward velocity, eventually halting it and causing the ball to fall back.

Time of Flight

Time of flight is the duration an object stays in the air. For objects thrown vertically, this is the sum of the time ascending and descending.
To compute it:

• Determine the time to reach maximum height using \( t = \frac{u}{a} \). For our ball, this was approximately \( 6.73 \, \text{seconds} \).
• Knowing the ascent and descent times are equal, multiply by two for total time. Here, this results in \( 13.46 \, \text{seconds} \).

Understanding time of flight is essential for predicting how long a projectile remains aloft, especially vital when calculating landing points or planning movements on different gravitational bodies like Mars.