Review Q #6 (Projectile Motion)

In summary, the track athlete had an initial speed of 10.2 m/s and made a jump of 8.8 meters. To calculate the most likely angle of take off, the equation R = [Vo^2 * sin2Ao] / g was used, with the known variables of range (8.8 m), initial velocity (10.2 m/s), and gravity (9.8 m/s^2). The maximum value of sin(2Ao) is 1, which occurs at an angle of 45 degrees. Therefore, the most likely angle of take off for the athlete was 45 degrees with respect to the horizontal.
  • #1
TeenieWeenie
30
0
Thanks Thaakisfox! Problem solved!

Homework Statement


A track athlete leaves the ground with initial speed 10.2 m/s and makes an 8.8 meter long jump. Calculate the most likely angle of the direction of his take off with respect to the horizontal.

Vo = 10.2 m/s
x = 8.8 m
xo = 0 m
Ao = ?

Homework Equations


x=Vo * CosAo * t
Vx = VoCosAo

The Attempt at a Solution


I have a missing t (time) variable or I'm using the wrong formula(s)?
 
Last edited:
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  • #2
Here you have to calculate a maximum. . Most probably the athlete jumped so that the distance was maximal. I would take the maximal height of the jump to be given and maximize with respect to the distance.
 
  • #3
R= [Vo^2 * sin2Ao] / g
where R = range = x = 8.8m

8 = [(10.2^2)*sin2Ao]/ 9.8m/s

8/9.8 = 10.2^2 *sin2Ao

[8/9.8]/(10.2^2) = sin2Ao

(sin2Ao = 2sinAocosAo ?)

[(8/9.8)/(10.2^2)]/2 = sinAoCosAo ?

Um...I don't know what to do from there...?

or if h was the given maximum height...
8 = [(10.2^2)sinAo] / 2(9.8)
[8/2(9.8)] = (10.2^2)sinAo
[8/19.6]/(10.2^2) = sinAo
sin-1[8/19.6]/(10.2^2) = Ao
.0039231 = Ao.

Errr...error?
 
  • #4
Yes. Thats the range of the projectile. So this has to be maximal. Now R is given, Vo is given, g is given, hence the only thing that can change is Ao the angle of the jump.
So we need sin(2Ao)= maximal. we know that the max of sin is 1. Hence sin(2Ao)=1.

Now at what angle does this occur?
 
  • #5
R= [Vo^2 * sin2Ao] / g
was the key to solving this problem then?
R = range of projectile = maximal = 8.
sin(2Ao) = maximal and will never be greater than 1 so it has to be = 1 which is 90 degrees or 2 * 45 degrees ?

Could you explain that a little more?
 
  • #6
Yes that's it. Exactly.
The athlete tries to jump, so that his/her (his in this case as female athletes usually don't jump above 7.5 :)) range is maximal. Now we have an expression for the range as a function of the velocity and the angle. We know that the jump was 8m, we also somehow know that the velocity was 10.2. Since we know that the range was maximal, and we know the velocity, we only have to figure out at what angle this could have happened.
And that is the sine should be 1. Which means that the angle of the jump is 45 degrees.
Its actually a good thing to remember that when you throw something (stone or anything messing around) the best is to throw it at a 45 degree angle with the horizontal because then it will go the furthest.
 
  • #7
OOO! I understand now!
That should have been a no-brainer :|

Thanks Thaakisfox! Problem solved!
 

1. What is projectile motion?

Projectile motion refers to the motion of an object that is launched into the air and then moves under the influence of gravity.

2. What factors affect projectile motion?

The factors that affect projectile motion include the initial velocity, the angle of launch, the force of gravity, and air resistance.

3. How do you calculate the range of a projectile?

The range of a projectile can be calculated using the formula R = (v^2 * sin(2θ)) / g, where v is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

4. What is the maximum height of a projectile?

The maximum height of a projectile can be calculated using the formula H = (v^2 * sin^2(θ)) / (2 * g), where v is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

5. How does air resistance affect projectile motion?

Air resistance can affect projectile motion by slowing down the object and altering its trajectory, leading to a shorter range and lower maximum height. However, for most practical purposes, air resistance can be ignored in projectile motion calculations.

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