What is the Average Velocity of a Stone Tossed in the Air?

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SUMMARY

The average velocity of a stone tossed in the air with an initial velocity of 20 m/s can be calculated using the height function h(t) = 20t - 4.9t². To find the average velocity over the time interval [1.5, 2.5], one must compute the heights at t = 1.5 s and t = 2.5 s, then apply the formula Vavg = change in distance/change in time. The discussion emphasizes that while the stone may reach high velocities, its average velocity over a complete round trip from ground level is zero due to the net distance traveled being zero.

PREREQUISITES
  • Understanding of kinematic equations, specifically height functions.
  • Familiarity with average velocity calculations.
  • Basic knowledge of quadratic functions and their properties.
  • Ability to perform arithmetic operations with real numbers.
NEXT STEPS
  • Calculate the height of the stone at t = 1.5 s and t = 2.5 s using the given height function.
  • Explore the concept of instantaneous velocity and how it differs from average velocity.
  • Learn about the effects of gravity on projectile motion in physics.
  • Investigate the implications of net displacement in calculating average velocity for different motion scenarios.
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Students studying physics, particularly those focusing on kinematics, as well as educators teaching concepts of motion and velocity.

vysero
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Here is my problem:

A stone is tossed in the air from ground level with an initial velocity of 20m/s. Its height at time t is h(t) = 20t-4.9t^2 m. Compute the stones avg velocity over the time interval [1.5,2.5].

I understand how to compute average velocity its Vavg=change in distance/change in time. However, I am running into a problem trying to figure a change in distance from the equation they gave me.

I tried searching in the library for a similar question but could not find one.
 
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What is the height at time t = 2.5 s?
What is the height at time t = 1.5 s?
How much time elapsed between time t = 1.5 s and time 2.5 s?
 
Oh I get it so I plug those values into the formula to get the height and then I compute the average velocity? Cool thanks!
 
Important point- if you throw a rock from ground level so that it goes to a very great height, then comes down and hits the ground again, obviously it must have had a large velocity at certain time. But since it started at ground level and got back to ground level, its net distance traveled is 0 and its average velocity over that time is 0.
 

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