Projectile and acceleration of gravity

[armoredfury16]

Homework Statement

During a baseball game, a batter hits a pop-
up to a fielder 86 m away.
The acceleration of gravity is 9.8 m/s2 .
If the ball remains in the air for 6.3 s, how
high does it rise? Answer in units of m.

Homework Equations

y=vi(sin theta)t-1/2g(t)^2

The Attempt at a Solution

I'm not even sure I went about this right...I was only able to fill in that equation this much:

y=vi (sin90)6.3-1/2(-9.8?)(6.3)^2

As you can see, I'm pretty clueless as to how to find the height of the ball at it's highest point...

 

[alphysicist]

Hi armoredfury16,

Homework Statement

During a baseball game, a batter hits a pop-
up to a fielder 86 m away.
The acceleration of gravity is 9.8 m/s2 .
If the ball remains in the air for 6.3 s, how
high does it rise? Answer in units of m.

Homework Equations

y=vi(sin theta)t-1/2g(t)^2

The Attempt at a Solution

I'm not even sure I went about this right...I was only able to fill in that equation this much:

y=vi (sin90)6.3-1/2(-9.8?)(6.3)^2

Since the minus sign has been pulled out in front of the 1/2, you don't need it again with the 9.8.

If you want to use that equation, you might try writing it twice--once for the ball reaching the ground, and one for the ball reaching the highest point. What do you get? (In particular, how long does it take for the ball to reach the highest point?)

 

[baba89]

I would like to first acknowledge the student's attempt at solving the problem using the given equation. However, there are a few errors in their attempt that I would like to address.

Firstly, the equation provided is correct, but the values for initial velocity and angle of launch (theta) are missing. In order to solve for the height, we need to know the initial velocity at which the ball was hit and the angle at which it was launched.

Secondly, the value of -9.8 m/s^2 should be used for the acceleration due to gravity, as it is a constant value on Earth.

To solve for the height, we can use the equation y = vi(sin theta)t - 1/2gt^2, where vi is the initial velocity, theta is the angle of launch, t is the time in seconds, and g is the acceleration due to gravity.

Using the given information, we can rearrange the equation to solve for the height (y). This would give us y = (86 m) + 1/2(9.8 m/s^2)(6.3 s)^2, which gives us a final height of approximately 100.38 m.

In conclusion, the ball rises to a height of approximately 100.38 m before falling back to the ground. It is important to note that this calculation assumes ideal conditions and does not take into account air resistance or other external factors.