How Do You Calculate the Initial Velocity of a Tennis Ball Thrown at an Angle?

AI Thread Summary
To calculate the initial velocity of a tennis ball thrown at a 30° angle, it is important to understand that velocity includes both magnitude and direction, while speed only refers to magnitude. The ball takes 1 second to reach its maximum height, indicating it must have an initial speed greater than zero, as it is moving upward against gravity. The initial vertical velocity can be calculated using the relationship between the vertical component and the overall velocity, with the vertical component being influenced by the angle of projection. The initial vertical speed at the peak is momentarily zero, and the calculation involves using the sine of the angle to find the total initial velocity. Understanding these concepts is crucial for solving projectile motion problems effectively.
Haniszmi
Messages
13
Reaction score
0

Homework Statement


A tennis ball is thrown from ground level with an initial velocity directed 30° above the horizontal. If it takes the ball 1 second to reach the top of its trajectory, what is the magnitude of the initial velocity?


Homework Equations





The Attempt at a Solution


I am new to physics and the only attempt I made towards this problem is a visual drawing of the problem. I do not understand what they mean by "what is the magnitude of the initial velocity?" Can someone please explain what I have to find here
 
Physics news on Phys.org
"what is the magnitude of the initial velocity?"

It simply means 'What is the initial speed?'

Velocity is a vector, which means it is a magnitude (size) AND a direction (30 degrees from horiz). Speed is a scalar, which means it only has magnitude, not direction.

You simply need to worry about the initial speed, and not the initial direction.
 
Ok... I understand what you told me and using what you told me my first gut reaction is to say that the answer is 0. Since the ball is not moving anywhere initially. Is this valid?
 
Haniszmi said:
Ok... I understand what you told me and using what you told me my first gut reaction is to say that the answer is 0. Since the ball is not moving anywhere initially. Is this valid?

No.

If it takes 1 second to reach the top then how high is that?
 
The ball has to have SOME initial speed greater than zero because it is traveling upwards. This is in the opposite direction to the acceleration due to gravity which is pulling the ball down and reducing the speed of the ball.

The maximum height that the ball reaches is when the ball has stopped traveling up and starts to fall back down because of gravity. Therefore, at time = 1 second, the ball's speed is momentarily zero.
 
Ok I have a feeling that the answer is 9.8m because once the ball is thrown in the air it automatically begins to decelerates because of gravity. Since it takes an object that was dropped to travel 9.8m in one second, an object that was thrown up and airborne for only one second must of been thrown up at an initial velocity of 9.8m/s as well because deceleration is -9.8m/s.
 
Haniszmi said:
Ok I have a feeling that the answer is 9.8m because once the ball is thrown in the air it automatically begins to decelerates because of gravity. Since it takes an object that was dropped to travel 9.8m in one second, an object that was thrown up and airborne for only one second must of been thrown up at an initial velocity of 9.8m/s as well because deceleration is -9.8m/s.

That is correct ... except that he threw it at an angle of 30 degrees.

But the good news is you have calculated the initial y velocity. And at 30 degrees. you can know what the original velocity is by the y component being sin 30 degrees of the original velocity.
 
Back
Top