91Ó°ÊÓ

In projectile motion, initial velocity is the speed at which an object is thrown or propelled from its starting point. For a vertically thrown object, this is crucial as it determines how high and how long the object will stay in motion.

In our problem, the faster stone is launched with three times the initial speed of the slower stone. The initial velocity can be found using the formula:

• \( v_{0} = g \cdot t_{max} \)
• Where \( g \) is the acceleration due to gravity (approximated as 9.8 m/s\(^2\)) and \( t_{max} \) is time to reach maximum height.

For the faster stone, with a time to maximum height of 5 seconds, the initial velocity is \( v_{0f} = 9.8 \times 5 = 49 \text{ m/s} \). For the slower stone, this initial speed is a third of the faster stone's, calculated as \( v_{0s} = \frac{49}{3} \approx 16.33 \text{ m/s} \). This initial velocity is foundational for determining other aspects of the stone's motion.

It sets the stage for calculating how long the stone will be in the air and the maximum height it will achieve.

The maximum height in projectile motion refers to the peak point in the stone's trajectory where the vertical velocity becomes zero momentarily before the stone starts its descent back to the ground.

This height is an important concept as it illustrates the impact of the initial velocity on reach. To calculate this for any thrown object, use:

• \( H = \frac{v_{0}^{2}}{2g} \)
• \( v_{0} \) is the initial velocity and \( g \) is the constant gravitational acceleration.

For the slower stone, substituting its velocity results in \( H_s = \frac{16.33^2}{2 \times 9.8} \approx 13.6 \text{ meters} \). The faster stone, with thrice the initial velocity, reaches much higher: \( H_f = \frac{49^2}{2 \times 9.8} = 122.5 \text{ meters} \).

This can also be expressed in terms of the slower stone's maximum height yield: \( H_f = 9 \times H_s \). Understanding maximum height helps us see how variations in initial velocity affect peak elevation.

The time of flight in projectile motion is the total duration from the launch until the object returns to its initial vertical position. It depends on the initial velocity and gravitational effects.

This concept helps us understand how varying speeds affect motion duration. For our problem:

• The faster stone's time of flight is given as 10 seconds, which includes ascent and descent.
• For the slower stone, calculate using its initial velocity \( v_{0s} \).

Using the formula \( v_{0} = g \times \frac{t_{flight}}{2} \), and solving, the time to peak, \( t_{max} \approx 1.67 \text{ seconds} \) leads to a total time of flight \( t_s \approx 3.34 \text{ seconds} \).

Knowing the time of flight is essential for understanding how different factors like initial velocity affect the duration of airborne movement in physics problems.

In projectile motion, acceleration due to gravity plays a crucial role as it influences how quickly an object slows down as it moves upward, and accelerates as it comes back down. This is a force that pulls objects toward the center of the Earth.

In calculations, gravity is denoted by \( g \) and approximately valued at 9.8 m/s\(^2\). It's a constant that affects the initial velocity, maximum height, and time of flight.

• It determines the rate at which an object moves upwards and reaches a velocity of zero at its peak.
• Then, gravity accelerates it downwards, thereby completing the projectile motion.

Gravity impacts how high and how long an object will travel after being projected or thrown. Understanding this helps discern the interplay between initial velocity and other motion aspects like maximum height and flight time, as seen in our stone throwing scenario.