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:

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