Review Q #5 (Projectile Motion)

In summary: The problem seems to be that you plugged in the incorrect cosine value. You should have used cosine of 45 degrees. After changing the equation to use the correct cosine value, the final answer comes out as 2.9698 m/s.
  • #1
TeenieWeenie
30
0

Homework Statement


A basketball leaves the hands of a shooter 2.1 meters above the floor and soars towards the basket, which is at a horizontal distance of 3 meter from the shooter. If the basket is 2.6 meters above the floor, and the player shoots the ball at an angle of 45 degree above the horizontal, calculate the initial velocity the shooter must give the basketball in order to score a basket.

Yo = 2.1m
Xo = 3m

Y = 2.6m
Ao = 45 degrees above horizontal (?)

Homework Equations


Vx = VoCosAo
Vy = VoSinAo - gt


The Attempt at a Solution


I don't understand what the X is supposed to be?
 
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  • #2
Ok, so I've drawn the diagram for this.
It looks like this is solved with trig?
I get a right triangle with the top as 2.6m. and the lower 45 degree angle as 2.1m.
So the height of the triangle would be 0.5m and the base would be 3m.

Then, what does the hypotenuse equal?
 
  • #3
This is projectile motion. No it cannot be solved with trig. You have to use physics too ;)
 
  • #4
Hehe.
What am I missing :( ?
I have no idea where to go from there (I've been looking at this for 10 minutes straight), yet again.
 
  • #5
First you know that the horizontal distance x=Vo*cos(45)*t.
And you know the vertical distance: y=Vo*sin(45)*t-(g/2)t^2.
Now from the first equation take t and plug it into the second equation. You will get an equation for y in terms of x.
You want that y=0.5 and x=3 right? as the total height difference between the point of throw and the height of the basket is (2.6-2.1=0.5).
 
  • #6
3 = Vo * cos45o * t
t = 3/ (Vo * cos45o)

0.5 = Vosin45o-(9.8/2)t2
0.5 = 3 tan45 - (9.8/2)(9/cos245oVo2

Vo2= [(3tan45o-0.5)/(9.8/2)]*(9/cos245o)

Vo2= (2.5/(9.8/2) * (0.5/9)
Vo2= 0.2834467
V = (0.2834467)^1/2 ?

That's...odd?
I don't think that's right...
 
  • #7
You messed up the calculation. Check that you expressed the velocity correctly.
 
  • #8
Alrighty, looks like I had computational error.

0.5 = 3 -(9.8*9/(Vo^2*cos^2(45))/2
11.025/ [Vo^2cos^2(45)]= 2.5
11.025/2.5 = Vo^2 * cos^2(45)
4.41 = [Vo^2]/2
8.82 = Vo^2
2.9698 m/s = Vo

how about that?
 
  • #9
That still looks wrong ;)
 
  • #10
What could be it!? :(
 
  • #11
Check from here again :
0.5 = 3 -(9.8*9/(Vo^2*cos^2(45))/2
and calculate step by step
 

What is projectile motion?

Projectile motion is the motion of an object through the air or space under the influence of gravity. It is a combination of horizontal and vertical motion, and the path it takes is a curved trajectory.

What are the factors that affect projectile motion?

The factors that affect projectile motion are the initial velocity, angle of projection, and the force of gravity. Air resistance can also affect the motion of a projectile.

How is the trajectory of a projectile calculated?

The trajectory of a projectile can be calculated using the equations of motion and the principles of trigonometry. The initial velocity, angle of projection, and gravitational force are used to determine the horizontal and vertical components of the projectile's motion, which can then be used to calculate its trajectory.

What is the difference between horizontal and vertical motion in projectile motion?

Horizontal motion is the motion of a projectile along the x-axis, while vertical motion is the motion along the y-axis. Horizontal motion is constant and unaffected by gravity, while vertical motion is affected by gravity and changes over time.

What are some real-life applications of projectile motion?

Some real-life applications of projectile motion include throwing a ball, shooting a basketball, firing a rocket, and launching a satellite into space. It is also used in sports, such as baseball, football, and golf, to determine the trajectory of a ball.

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