Projectile Motion (No initial velocity)

In summary, to reach a waterfall that is 0.55 m tall, a salmon starting at 2.00 m away must have a minimum initial speed that can be found by using the formula \Deltay=visin\Theta\Deltat+1/2g\Deltat2 and writing expressions for vertical and horizontal displacement at any given time.
  • #1
th7172
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Homework Statement


A salmon starts 2.00 m from a waterfall that is 0.55 m tall and jumps at an angle of 32.0. What must be the salmon's minimum initial speed to reach the waterfall.


Homework Equations


[tex]\Delta[/tex]y=visin[tex]\Theta[/tex][tex]\Delta[/tex]t+1/2g[tex]\Delta[/tex]t2

The Attempt at a Solution


I have tried all the related formulas for projectile motion, and I felt like I got somewhere a few times. But the results aren't really checking with a graphing calculator.
 
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  • #2
Try to write expressions for the vertical and horizontal displacement at any given time (in terms of the unknown initial speed, ofcourse).

For some particular value of time, the vertical displacement will be 0.55m, and at the same time, the horizontal displacement will be 2m. You can use this information to find v.
 
  • #3


I can understand your frustration in trying to solve this problem and getting different results. Projectile motion is a complex concept and can be difficult to solve without the proper understanding of the equations and principles involved. It is important to first understand the basics of projectile motion, such as the horizontal and vertical components of motion, the acceleration due to gravity, and the relationship between time and distance.

To solve this problem, you can use the equation \Deltay=visin\Theta\Deltat+1/2g\Deltat2, where \Deltay is the vertical displacement, vi is the initial velocity, \Theta is the angle of the jump, g is the acceleration due to gravity, and \Deltat is the time.

Since the salmon starts 2.00 m from the waterfall and the waterfall is 0.55 m tall, the vertical displacement is 2.55 m. The angle of the jump is given as 32.0, and we know that the acceleration due to gravity is -9.8 m/s^2 (negative because it acts in the opposite direction of the jump). We can rearrange the equation to solve for vi:

\Deltay=visin\Theta\Deltat+1/2g\Deltat2
2.55 = vi*sin(32)*\Deltat + 1/2(-9.8)*\Deltat^2
Solving for \Deltat, we get:
\Deltat = 2.55/(vi*sin(32))

Substituting this value of \Deltat into the equation and rearranging, we get:
vi = 2.55*sin(32)/\Deltat

Now, we need to find the minimum initial velocity that will allow the salmon to reach the waterfall. This will happen when the time taken to reach the waterfall is the minimum possible, which is when the vertical displacement is equal to the height of the waterfall (0.55 m). So, we can set \Deltay = 0.55 and solve for vi:

0.55 = vi*sin(32)*\Deltat + 1/2(-9.8)*\Deltat^2
Solving for \Deltat, we get:
\Deltat = 0.55/(vi*sin(32))

Substituting this value of \Deltat into the equation and rearranging,
 

1. What is projectile motion?

Projectile motion is the motion of an object that is launched into the air at an angle, and then moves along a predetermined path due to the force of gravity.

2. What is the formula for calculating projectile motion?

The formula for calculating projectile motion is: x = v0 * t * cosθ, where x is the horizontal displacement, v0 is the initial velocity, t is the time, and θ is the angle of launch.

3. How does the angle of launch affect projectile motion?

The angle of launch affects the horizontal displacement and the maximum height reached by the object. A higher angle of launch will result in a longer horizontal displacement, while a lower angle will result in a shorter horizontal displacement but a higher maximum height.

4. Does air resistance affect projectile motion?

Yes, air resistance can affect projectile motion by slowing down the object as it travels through the air. However, for most practical purposes, the effect of air resistance is negligible.

5. What is the trajectory of a projectile in the absence of initial velocity?

If there is no initial velocity, the projectile will follow a parabolic path due to the force of gravity. The trajectory will be a perfect semicircle, with the object reaching its highest point at the midpoint of its path and then falling back down to its initial height.

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