Basketball Projectile Motion Problem

[Jeff231]

Homework Statement

A basketball player who is 2.00[m] tall is standing on the floor 10.0 [m] from the basket, as in the figure. If he shoots the ball at 40.0degrees angle with the horizontal, at what initial speed must he throw so that it goes through the hoob without striking the backboard? The basket height is 3.05[m].

Homework Equations

xf = xi + vit + 1/2at^2
yf= yi +vit + 1/2at^2

The Attempt at a Solution

I used the yf equation to come up with an equation that has vi and t as the variables. Then I used the xf equation and solved for t. Then I plugged in that t into the yf equation. It comes out to a pretty large quadratic equation:

-4.9 (10/vi(sin40))^2 + (vi(cos40))(10/vi(sin40)) - 1.05 = 0

Now I'm having trouble finding vi. I'm pretty sure I did everything right, but can someone help me figure out how to solve for vi?

Also, how do you make it so the quations come up looking normal instead of typing it here, it's formatted different? Is that a website you type it in and link here? Thanks!

-Jeff

 

[LowlyPion]

Homework Statement

A basketball player who is 2.00[m] tall is standing on the floor 10.0 [m] from the basket, as in the figure. If he shoots the ball at 40.0degrees angle with the horizontal, at what initial speed must he throw so that it goes through the hoob without striking the backboard? The basket height is 3.05[m].

Homework Equations

x_f = x_i + v_it + 1/2at²
y_f = y_i + v_it + 1/2at²

The Attempt at a Solution

I used the yf equation to come up with an equation that has vi and t as the variables. Then I used the xf equation and solved for t. Then I plugged in that t into the yf equation. It comes out to a pretty large quadratic equation:

-4.9 (10/vi(sin40))^2 + (vi(cos40))(10/vi(sin40)) - 1.05 = 0

Now I'm having trouble finding vi. I'm pretty sure I did everything right, but can someone help me figure out how to solve for vi?

Also, how do you make it so the quations come up looking normal instead of typing it here, it's formatted different? Is that a website you type it in and link here? Thanks!

-Jeff

First of all I hope you didn't show any acceleration in the x direction. I haven't checked what you did, but that term is 0 for V_x.

As to figuring combined x,y speed remember that the x and y components of speed add like vectors.

As to the superscripting and subscripting, I modified your Relevant Equations to show you how to code it. You can also use TEX (LaTex) but that is a more complicated tutorial. If you open up some of the posts with formulas, if you can make a 10 m jumpshot, that won't be so hard.

 

[baba89]

Dear Jeff,

Your approach to solving this problem is correct. However, there are a few steps missing in your solution that may be causing difficulty in finding the initial velocity (vi). Let me guide you through the steps to find the solution.

Step 1: Drawing a diagram
It is always helpful to draw a diagram of the problem to visualize the situation. In this case, you can draw a diagram showing the basketball player standing 10.0m away from the basket, with an angle of 40 degrees and a height of 2.00m.

Step 2: Identifying the known and unknown variables
From the given information, we can identify the following variables:
- Initial vertical position (yi) = 2.00m
- Final vertical position (yf) = 3.05m
- Initial horizontal position (xi) = 10.0m
- Angle with the horizontal (θ) = 40 degrees
- Acceleration due to gravity (a) = -9.8m/s^2 (negative sign indicates downward direction)
- Initial velocity (vi) = unknown
- Time taken (t) = unknown

Step 3: Writing the equations
Using the equations of motion, we can write the following equations:
xf = xi + vit + 1/2at^2
yf = yi + vit + 1/2at^2

Step 4: Solving for t
Since we are trying to find the initial velocity (vi), we need to eliminate the variable t. To do this, we can solve for t in the xf equation and substitute it into the yf equation. The xf equation becomes:
t = (xf - xi) / (vi cos θ)
Substituting this value of t into the yf equation, we get:
yf = yi + (vi sin θ)(xf - xi) / (vi cos θ) + 1/2a((xf - xi) / (vi cos θ))^2

Step 5: Simplifying the equation
Now, we can simplify the equation by multiplying both sides by (vi cos θ) and rearranging terms:
0 = 1/2a(vi cos θ)^2 - (vi sin θ)(xf - xi) + (yf - yi)(vi cos θ)

Step 6: Solving for vi
This equation is now in the form