Projectile Motion Problem: Determining Height Using Equations

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SUMMARY

The projectile motion problem involves calculating the height of a leaf above water level, given that a fish shoots water at a 60-degree angle with an initial velocity of 3.01 m/s. The horizontal distance to the leaf is 0.600 m. The correct approach requires determining the time for the water to travel horizontally, which is calculated as 0.398 seconds. The vertical position of the water at this time is found using the equation for vertical motion, yielding a height of 0.259 m. Rounding intermediate values can lead to inaccuracies, so it is essential to use exact values throughout the calculations.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with trigonometric functions (sine and cosine)
  • Knowledge of kinematic equations for motion
  • Basic calculator usage for precise calculations
NEXT STEPS
  • Study the derivation and application of kinematic equations in projectile motion
  • Learn how to accurately use trigonometric functions in physics problems
  • Explore the effects of rounding errors in calculations and how to minimize them
  • Practice similar projectile motion problems with varying angles and velocities
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators looking for examples of problem-solving in kinematics.

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Homework Statement



An insect is resting above the water level on a leaf, a horizontal 0.600 m away from a fish. The fish shoots water into the air above the surface of the water at a 60 degree angle, with an initial velocity of 3.01 m/s.

Assume the fish shoots the water into the air at water level, and that the water hits the insect.

At what height is the leaf above water level?

Homework Equations


v = vi + at
v= d/t
x = vi(t) + 0.5a(t^2)

The Attempt at a Solution


First I tried to find out how much time it would take the water to travel in the horizontal direction.

Initial velocity in horizontal direction: cos60*3.01= 1.505 m/s
With zero acceleration, the time it would take to travel 0.600 m would be:
t=0.600/1.505=0.398 s

Then, I tried to figure out where the water would be at 0.398 seconds vertically.
Its initial velocity is sin60*3.01 = 2.60 m/s
2.60(0.398)+(0.5)(-9.81)(0.398^2)= 0.259 m

I must be doing something wrong because the program says I have the wrong answer. I have no idea where the problem is, though... any ideas?
 
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You have rounded at the intermediate steps, which has induced errors in your final answer. Try to use the exact values of the intermediate steps by either storing them in your calculator, or writing down the full calculator display.
 
Thanks, I think that solves the problem.
 

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