Physics 20-1 Final Review: Projectiles

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SUMMARY

The discussion focuses on solving a projectile motion problem involving an object thrown from a 75 m tall building with an initial speed of 16 m/s at a 25-degree angle. Key calculations include determining the time to reach maximum height using the vertical component of velocity (16*sin25) and the subsequent free fall to the ground. The correct approach involves breaking the motion into horizontal and vertical components, applying the equations of motion, and calculating the horizontal distance traveled using the total time of flight. The final distance from the base of the building is derived from these calculations.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with trigonometric functions (sine and cosine)
  • Knowledge of kinematic equations for motion (e.g., d = vit + 1/2at²)
  • Ability to analyze motion in two dimensions
NEXT STEPS
  • Calculate the vertical component of initial velocity using 16*sin25
  • Determine the time to maximum height with the formula t = (16*sin25)/g
  • Apply the free fall equation to find the time to hit the ground
  • Use the horizontal component of velocity (16*cos25) to calculate the total horizontal distance
USEFUL FOR

Students studying physics, particularly those preparing for exams in projectile motion, as well as educators looking for problem-solving strategies in kinematics.

physicznoob
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Homework Statement



A student is standing on the top of a building and he throws an object into the air with a speed of 16 m/s at an angle of 25 degrees above the horizontal. If the building is 75 m tall, how far from the base of the building will the object hit the ground?

Homework Equations



v = d/t
d = vit + 1/2at2

The Attempt at a Solution



i tried to break it up into horizontal and vertical components and break it down into speed, then distance, then time. but nothing i do seems to work.
 
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What do you get for the time to max height?
 
a = v2 - v1 / t
t = v2 - v1/ a
t = 0 - 16 / -9.81
t = 1.63 s.
 
There's your problem right there.

The vertical component of velocity is what must be taken to determine time to max height.

That is 16*sin25.

The horizontal component is 16*cos25.
 
i still get the wrong answer...
 
where are u setting your zero? the top of the building or the bottom?
 
huh? if you mean velocity wise, at the top for vertical motion.
 
physicznoob said:
i still get the wrong answer...

So what is your time to max height?

You know that must be 16*Sin25/g = t

And how high is that from launch? y = 1/2 g*t2

Now it is a simple free fall problem to determine the time to hit the ground isn't it?

With the total time ... and the horizontal component of velocity ... viola your distance along the ground.
 

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