How Do You Solve a Projectile Motion Problem?

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The discussion focuses on solving a projectile motion problem involving a projectile fired at an initial speed of 46.6 m/s and an angle of 42.2 degrees. Key calculations include determining the maximum height, total time in the air, and horizontal distance covered, with results showing a maximum height of 48.86 m, total air time of 6.314 seconds, and a range of 194.08 m. There is confusion regarding the symbols used in the equations and whether the initial speed is correctly represented. Additionally, questions arise about the calculations for the projectile's speed and direction 1.50 seconds after firing. Clarification on these points is necessary for accurate problem-solving.
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Homework Statement


A projectile is fired with an initial speed of 46.6 m/s at an angle of 42.2 ∘ above the horizontal on a long flat firing range.

1-Determine the maximum height reached by the projectile.

2-Determine the total time in the air.

3-Determine the total horizontal distance covered (that is, the range).

4-Determine the speed of the projectile 1.50 s after firing.

5-Determine the direction of the motion of the projectile 1.50 s after firing.

Homework Equations


v = vo +at x = xo + volt + .5at2 v2 = vo2 + 2a(x - xo) x=vt

The Attempt at a Solution


A.) Determine the maximum height Ymax reached by the projectile.

Ymax = (V*sin 45.2)^2/2g = (43.6*0.71)^2/19.612 = 48.86 mB.)Determine the total time t in the air.

t = 2*V*sin 45.2/g = 2*43.6*0.71/9.806 = 6.314 secC.)Determine the total horizontal distance covered (that is, the range).

Δx = V*cos 45.2*t = 43.6*0.705*6.314 = 194.08 m
 
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blackcemre said:

Homework Statement


A projectile is fired with an initial speed of 46.6 m/s at an angle of 42.2 ∘ above the horizontal on a long flat firing range.

1-Determine the maximum height reached by the projectile.

2-Determine the total time in the air.

3-Determine the total horizontal distance covered (that is, the range).

4-Determine the speed of the projectile 1.50 s after firing.

5-Determine the direction of the motion of the projectile 1.50 s after firing.

Homework Equations


v = vo +at x = xo + volt + .5at2 v2 = vo2 + 2a(x - xo) x=vt

The Attempt at a Solution


A.) Determine the maximum height Ymax reached by the projectile.

Ymax = (V*sin 45.2)^2/2g = (43.6*0.71)^2/19.612 = 48.86 mB.)Determine the total time t in the air.

t = 2*V*sin 45.2/g = 2*43.6*0.71/9.806 = 6.314 secC.)Determine the total horizontal distance covered (that is, the range).

Δx = V*cos 45.2*t = 43.6*0.705*6.314 = 194.08 m

So what's the question? You couldn't do #4 and #5?

Zz.
 
blackcemre said:
A projectile is fired with an initial speed of 46.6 m/s at an angle of 42.2 ∘ above the horizontal

blackcemre said:
A.) Determine the maximum height Ymax reached by the projectile.

Ymax = (V*sin 45.2)^2/2g = (43.6*0.71)^2/19.612 = 48.86 m
The numbers you've used in your work do not match those stated in the problem statement. Which numbers are correct?
 
There is confusion about the symbols used.
Is V the initial speed?
In the relevan equations, there are two different equations for x.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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