Difficult projectile motion question

In summary: Since the projectile starts 1m above the ground y=1 when x=01 = -g/2v²cos²αand when the projectile strikes the ground y=0 and x=600 = 60tanα - g/2v²cos²αfrom these two equations you can find tanα and plug that into1 = -g/2v²cos²αto solve for v.
  • #1
sorax123
34
0

Homework Statement



A cricketer strikes the ball at a height of 1 metre. It passes over a fielder 7 metres from the ball at a height of 3 metres, and hits the ground 60 metres from the bat. How fast was the ball hit?

The issue I have here is that after forming a quadratic in terms of tanα, you have u in as an unknown and subbing it back in gives incredibly messy algebra, which makes me think it's not the best way.

Homework Equations



u in x direction= ucosα s in x direction= utcosα
u in y direction= usinα s in y direction= utsinα - 4.9t^2

The Attempt at a Solution



My solution attempt is pages long and contains stupidly complex algebra, I think I am fundamentally missing something as the maths seems too fiddly for a fairly straightforward mechanics course.
Any tips to make the problem easier, or preferably a solution would be great. Thanks
 
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  • #2
You have probably made an invalid assumption or a mistake in your algebra or arithmetic. Without looking at your calculations, it is hard to point to the specific cause of your difficulty.
 
  • #3
Using the horizontal and vertical components of initial velocity, not the angle, the solution seems straightforward.
 
  • #4
This is the important formula to know:

##y = y_0 + tan{θ} x - \frac{g}{2 v_0^2 cos^2{θ}} x^2##

It has the form

##y = c + bx + ax^2##

and the problem gives us 3 points, so apply Gaussian elimination (sorry, GE is not something you will have learned yet, you need to solve 3 equations in 3 unknowns) to find a,b,c.

PS. I think this formula is worth memorizing. You should be able to see where it comes from.
 
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  • #5
sorax123 said:

Homework Statement



A cricketer strikes the ball at a height of 1 metre. It passes over a fielder 7 metres from the ball at a height of 3 metres, and hits the ground 60 metres from the bat. How fast was the ball hit?

The issue I have here is that after forming a quadratic in terms of tanα, you have u in as an unknown and subbing it back in gives incredibly messy algebra, which makes me think it's not the best way.

Homework Equations



u in x direction= ucosα s in x direction= utcosα
u in y direction= usinα s in y direction= utsinα - 4.9t^2

The Attempt at a Solution





My solution attempt is pages long and contains stupidly complex algebra, I think I am fundamentally missing something as the maths seems too fiddly for a fairly straightforward mechanics course.
Any tips to make the problem easier, or preferably a solution would be great. Thanks

If while passing over the fielder the projectile was at its max. Height it would make the problem more appropriate for the course level you describe. Did the problem perhaps say this? If the problem did state this you could solve for time in the air at that point and then everything becomes very straightforward.
 
  • #6
pgardn said:
If while passing over the fielder the projectile was at its max. Height it would make the problem more appropriate for the course level you describe. Did the problem perhaps say this? If the problem did state this you could solve for time in the air at that point and then everything becomes very straightforward.

Probably not a good assumption considering that the trajectory (a section of a parabola) must by symmetric about the point of maximum height. The fielder is only 7m from the batter. The ball lands 60m away.

One way to reduce the math burden might be to leave the initial vertical and horizontal components as separate variables, then write equations for the separate known points that the ball passes through. Hint: Lot's of nasty stuff should cancel cleanly if you translate the heights so that the bat is at height 0 (so, for example, the ball hits the ground at a "height" of -1m). Equate expressions for Vy to find Vx. This is one of those occasions where it's probably beneficial to substitute in the given numbers just before the end, allowing you to collapse a lot of terms before you hit the quadratic formula.
 
  • #7
gneill said:
Probably not a good assumption considering that the trajectory (a section of a parabola) must by symmetric about the point of maximum height. The fielder is only 7m from the batter. The ball lands 60m away.

One way to reduce the math burden might be to leave the initial vertical and horizontal components as separate variables, then write equations for the separate known points that the ball passes through. Hint: Lot's of nasty stuff should cancel cleanly if you translate the heights so that the bat is at height 0 (so, for example, the ball hits the ground at a "height" of -1m). Equate expressions for Vy to find Vx. This is one of those occasions where it's probably beneficial to substitute in the given numbers just before the end, allowing you to collapse a lot of terms before you hit the quadratic formula.
Yep.

My additional information makes no sense. And since they give them 3 points in a parabola...
It's a math problem.
 
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  • #8
gneill said:
One way to reduce the math burden might be to leave the initial vertical and horizontal components as separate variables, then write equations for the separate known points that the ball passes through. Hint: Lot's of nasty stuff should cancel cleanly if you translate the heights so that the bat is at height 0 (so, for example, the ball hits the ground at a "height" of -1m). Equate expressions for Vy to find Vx.

I would use the velocity components and exclude time, thus getting an equation for x and y. Then having a non-zero initial height should not complicate much of anything.
 
  • #9
You can rewrite the general equation of the projectile
y = yo + x tanθ - 1/2*g*x2/v2cos2θ as

1/2*g*x2/v2cos2θ = yo + x tanθ - y.
Write down two equations for two values of x and y. By making the division, you can simplify the equations and then solve for tanθ.
 
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1. What is projectile motion?

Projectile motion is the motion of an object that is projected or thrown into the air at an angle and moves along a curved path due to the influence of gravity.

2. What makes a projectile motion question difficult?

A difficult projectile motion question typically involves multiple variables, such as initial velocity, angle of projection, and air resistance, and requires the application of complex mathematical equations and concepts.

3. How do you solve a difficult projectile motion question?

To solve a difficult projectile motion question, you must first identify the given variables and understand the problem. Then, you can use equations such as the kinematic equations and the equations of motion to calculate the unknown variables and solve the problem step by step.

4. What are some common mistakes when solving projectile motion problems?

Some common mistakes when solving projectile motion problems include forgetting to account for air resistance, using the wrong formula or equation, and not correctly converting units of measurement.

5. How can understanding projectile motion be useful in real life?

Understanding projectile motion can be useful in various real-life situations, such as sports, engineering, and physics. For example, understanding the trajectory of a ball in sports can help improve performance, and understanding the motion of a projectile in engineering can aid in the design of structures and machines.

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