Throwing a ball and calculating the speed

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In summary: It seems that you have not done anything incorrectly. All you are asked to do is calculate the speed of the ball at the start of its free flight, given the time of free flight, distance travelled, etc.
  • #1

Homework Statement


A person with a 3 kg ball at rest in his hand, throws it with constant force at an angle of 45 degrees. The ball moves in a straight line until it leaves his hand at 2.1 meters above the ground. His arm has the length of 0.55 meters. Between the ball leaving his hand and until impact the total time elapsed is 2.15 seconds and the horizontal distance is 16 meters.

What is the speed of the ball the moment it leaves his hand?

Homework Equations


Constant acceleration on a straight line:
[itex]v_x=v_{0x}+a_xt[/itex]

[itex]x=x_{0}+v_{0x}t+\frac{1}{2}a_{x}t^2[/itex]

[itex]v^2_x=v^2_{0x}+2a_x(x-x_0)[/itex]

[itex]x-x_0=(\frac{v_{0x}+v_x}{2})t[/itex]

Projectile motion:
[itex]x=(v_0 \cos(\alpha_0))t[/itex]

[itex]y=(v_0 \sin(\alpha_0))t-\frac{1}{2}gt^2[/itex]

[itex]v_x=v_0 \cos(\alpha_0)[/itex]

[itex]v_y=v_0 \sin(\alpha_0)-gt[/itex]

The Attempt at a Solution


The text tells me that from the ball is at rest and until it leaves his hand, the movement is a straight line. Therefore I'm thinking I should use one of the equation for straight line.
Also, if the force is constant it means the acceleration is constant as well (can somebody please confirm this to me?)

So I want to find [itex]v_x[/itex]. I rearrange my coordinate system so that the straight line movement his arm makes is parallel to the x-axis.

Thus the distance that the ball travels in this straight line is the length of the guy's arm, which is 0.55 meters so:
[itex]x-x_0=0.55 m[/itex]

I also know that since the ball is initially at rest:
[itex]v_{0x}=0[/itex]

But I don't know the acceleration during the straight line motion nor do I know the time it takes for the ball from rest until it leaves his hand. Therefore I have no idea which equation to use.
Maybe I should be looking at the projectile motion equations instead?
 
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  • #2
It seems that you are given a lot more information here than is required to determine the answer. All you are asked to find is the speed of the ball at the start of its free flight: given the time of free flight, distance travelled, etc., find its initial speed. What happens prior to the moment it leaves his hand seems irrelevant.
 
  • #3
NascentOxygen said:
It seems that you are given a lot more information here than is required to determine the answer. All you are asked to find is the speed of the ball at the start of its free flight: given the time of free flight, distance travelled, etc., find its initial speed. What happens prior to the moment it leaves his hand seems irrelevant.

There are more questions to the text but I just included everything to be on the safe side. :)

So are you saying this is the answer:
[itex]16m=(v_0 \cos(45deg))*2.15s[/itex]
[itex]v_0=10.5m/s[/itex]

But the thing is that the ball is in projectile motion 2.1 meters above the ground and then lands 16 meters on the ground, so shouldn't I account for those 2.1 meters in some way?
 
  • #4
PhyIsOhSoHard said:
There are more questions to the text but I just included everything to be on the safe side. :)

So are you saying this is the answer:
[itex]16m=(v_0 \cos(45deg))*2.15s[/itex]
[itex]v_0=10.5m/s[/itex]
Looks right.
But the thing is that the ball is in projectile motion 2.1 meters above the ground and then lands 16 meters on the ground, so shouldn't I account for those 2.1 meters in some way?
No. If an object has a horizontal component of velocity v for time t then the horizontal distance moved is vt. What's happening in the vertical at the same time is irrelevant. You next need to use the time calculated and the 2.1m to compute the vertical component of the launch velocity.
 
  • #5
haruspex said:
Looks right.

No. If an object has a horizontal component of velocity v for time t then the horizontal distance moved is vt. What's happening in the vertical at the same time is irrelevant. You next need to use the time calculated and the 2.1m to compute the vertical component of the launch velocity.

I just find it strange that I'm not using the initial height of the ball and the length of the person's arm for anything.

I calculated the highest point that the ball reaches in motion.
The time:
[itex]v_y=0[/itex]

[itex]0=10.5m/s \sin(45deg)-9.8m/s^2*t[/itex]
[itex]t=0.76s[/itex]

The height:
[itex]y=(10.5m/s \sin(45deg))*0.76s-\frac{1}{2}*9.8m/s^2*(0.76s)^2[/itex]
[itex]y=2.8m[/itex]

I'm also supposed to calculate the speed when it hits the ground. I have been given in the text that it hits the ground at 2.15 s

I calculate the x and y components for the velocity:
[itex]v_x=10.5m/s \cos(45deg)=7.4m/s[/itex]
[itex]v_y=10.5m/s \sin(45deg)-9.8m/s^2*2.15s=-13.7m/s[/itex]

But again I didn't use the 2.1 m height or the 0.55 m length of his arm.
Have I done something wrong?
 
  • #6
PhyIsOhSoHard said:
I just find it strange that I'm not using the initial height of the ball and the length of the person's arm for anything.
The length of arm, and all info pertaining to the time before release, must be for later questions. But you are right that there is soemthing wrong with the question. You have too much data on what happens after release, and it conflicts. From the given flight time and horizontal distance the launch speed must be about 7.5m/s horizontally. Since it's at 45 deg that's also the vertical speed. The vertical displacement in that time is the horizontal displacement less gt2/2, or about 16-20 =-4 m. That doesn't fit with the -2.1 m given. maybe this is in an environment where g has a different value.
 
  • #7
haruspex said:
The length of arm, and all info pertaining to the time before release, must be for later questions. But you are right that there is soemthing wrong with the question. You have too much data on what happens after release, and it conflicts. From the given flight time and horizontal distance the launch speed must be about 7.5m/s horizontally. Since it's at 45 deg that's also the vertical speed. The vertical displacement in that time is the horizontal displacement less gt2/2, or about 16-20 =-4 m. That doesn't fit with the -2.1 m given. maybe this is in an environment where g has a different value.

I have a last question that asks me to draw a free body diagram during the throw. Perhaps that's where the length of his arm and the height when the ball leaves his hand comes into play, right?
 
  • #8
PhyIsOhSoHard said:
I have a last question that asks me to draw a free body diagram during the throw. Perhaps that's where the length of his arm and the height when the ball leaves his hand comes into play, right?
The length of the arm, yes, but not the launch height. Because the question provides conflicting data for the trajectory after launch you can get different answers according to which subset of data you use.
 
  • #9
haruspex said:
The length of the arm, yes, but not the launch height. Because the question provides conflicting data for the trajectory after launch you can get different answers according to which subset of data you use.

But my methods for finding the initial speed, the highest point in the trajectory and the speed when it hits the ground, were all correct right? regardless of the answers?
 
  • #10
PhyIsOhSoHard said:
But my methods for finding the initial speed, the highest point in the trajectory and the speed when it hits the ground, were all correct right? regardless of the answers?
Yes.
 
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1. How do you calculate the speed of a thrown ball?

The speed of a thrown ball can be calculated by dividing the distance the ball traveled by the time it took to travel that distance. This is known as the average speed formula, which is speed = distance / time.

2. What is the difference between speed and velocity when throwing a ball?

Speed is a scalar quantity that only measures the magnitude of an object's motion, while velocity is a vector quantity that measures both the magnitude and direction of an object's motion. When throwing a ball, the speed would be the numerical value of how fast the ball is traveling, while the velocity would include the direction the ball is being thrown.

3. How do you measure the distance a ball has traveled when thrown?

The distance a ball has traveled when thrown can be measured by using a measuring tool, such as a tape measure or ruler. The starting point would be where the ball was thrown from, and the end point would be where the ball lands. The distance between these two points is the distance the ball has traveled.

4. What factors can affect the speed of a thrown ball?

There are several factors that can affect the speed of a thrown ball, including the force used to throw the ball, the angle at which the ball is thrown, the weight and shape of the ball, and any external factors such as air resistance or wind.

5. Can the speed of a thrown ball be calculated without measuring the distance it has traveled?

No, the speed of a thrown ball cannot be accurately calculated without measuring the distance it has traveled. The average speed formula requires both distance and time to be known in order to calculate the speed. Without measuring the distance, there would be no way to accurately determine the speed of the thrown ball.

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