How Do You Calculate Projectile Motion and Vertical Displacement?

[Coco12]

Homework Statement

A baseball player leads off the game and hits a long ball to right field. The ball leaves the bat at an
angle of 30.0o from the horizontal with a horizontal velocity of 40.0 m/s. How far will it travel in the air?

2) person throws bottle straight up in the air at velocity of 15m/s . How high is the bottle after 2 secs

Homework Equations

Vy=Vyi +gt
Vy=Vyit +1/2gt^2

The Attempt at a Solution

I found the time in total to be 4.08. However this is asking time IN the air. Isn't that different from finding Vxi and multiplying by time ? How do I find it?

2)can I use by y=15m/s* 2+1/2* 9.8m/s^2 * 2
Is that correct?

 

[gneill]

Homework Statement

A baseball player leads off the game and hits a long ball to right field. The ball leaves the bat at an
angle of 30.0o from the horizontal with a horizontal velocity of 40.0 m/s. How far will it travel in the air?

2) person throws bottle straight up in the air at velocity of 15m/s . How high is the bottle after 2 secs

Homework Equations

Vy=Vyi +gt
Vy=Vyit +1/2gt^2

The Attempt at a Solution

I found the time in total to be 4.08. However this is asking time IN the air. Isn't that different from finding Vxi and multiplying by time ? How do I find it?

2)can I use by y=15m/s* 2+1/2* 9.8m/s^2 * 2
Is that correct?

For (1), can you give details on how you found the time? Your value doesn't look quite right to me. I take it from the level of question that "how far it travels in the air" is referring to the horizontal distance traveled between leaving the bat and first hitting the ground.

For (2), you need to take into account the direction of motion versus acceleration. Is the acceleration due to gravity directed upwards or downwards. Also, don't forget to square the time on the acceleration term! Otherwise, you've chosen the right form of equation.

 

[Coco12]

So for 2 I would be taking 15*2+ (.5* -9.8m/s2 * 2^2) right?

 

[gneill]

So for 2 I would be taking 15*2+ (.5* -9.8m/s2 * 2^2) right?

Right. Or symbolically:

$y = v_o t - \frac{1}{2}g t^2$

 

[Nickie]

I would first clarify any uncertainties or missing information in the problem. For the first problem, I would ask whether the height of the right field fence or any other obstacles need to be taken into account when calculating the distance the ball will travel. I would also confirm the units for the horizontal velocity and clarify whether the angle given is the launch angle or the angle of the ball's trajectory at any given point.

For the second problem, I would clarify whether the bottle is thrown from ground level or from a certain height. I would also ask for the units of time and confirm whether the acceleration due to gravity is -9.8 m/s^2.

Once all the necessary information is gathered, I would use the equations of projectile motion to solve the problems. For the first problem, I would use the equation d = v*t*cos(theta) to find the horizontal distance the ball travels, where d is distance, v is velocity, t is time, and theta is the launch angle. For the second problem, I would use the equation h = v*t + 1/2*g*t^2, where h is the height, v is the initial velocity, g is the acceleration due to gravity, and t is time.

It is important to note that the equations used may vary depending on the specific circumstances of the problem, such as the presence of air resistance. I would also consider any potential sources of error in the calculations and make sure to use accurate and precise values for all variables.