Can a Baseball Hit at 26 Degrees Clear a 3.8m Fence 120m Away?

[moondawg]

QUESTION: A baseball is hit 1.2 meters above the ground, at an angle of 26 degrees, with an initial speed of 80 m/s.
a. Will it clear a 3.8m fence 120m away from the home-run?

MY PROCESS(But not positive I am doing it correctly)
KNOWN
V_x= 80cos26=71.9 m/s
V_y= 80sin26=35.07 m/s
So I did the problem as if from ground level and tried to find time.(using -10 as my acceleration thankyou Mr Werner)
> 0=35.07+-10t t=3.2x2=7s
*got my time to be 7s
Then I am completely lost on what to do after that. I tried finding the how high the ball is when it is insantaneously at 120m in the x direction but my numbers are so extremely wrong... HELLLP MEEEEEE! please

 

[alphysicist]

Hi moondawg,

QUESTION: A baseball is hit 1.2 meters above the ground, at an angle of 26 degrees, with an initial speed of 80 m/s.
a. Will it clear a 3.8m fence 120m away from the home-run?

MY PROCESS(But not positive I am doing it correctly)
KNOWN
V_x= 80cos26=71.9 m/s
V_y= 80sin26=35.07 m/s
So I did the problem as if from ground level and tried to find time.(using -10 as my acceleration thankyou Mr Werner)
> 0=35.07+-10t t=3.2x2=7s
*got my time to be 7s

You solved this equation, and then doubled the time to find the time of 7s. But what exactly does this particular time represent? In other words, what is the ball doing (or what can you say about its position) at t=7s?

(Remember that the goal of this problem is to find out if the ball makes is over the fence or not.)

Then I am completely lost on what to do after that. I tried finding the how high the ball is when it is insantaneously at 120m in the x direction but my numbers are so extremely wrong... HELLLP MEEEEEE! please

 

[jmb88korean]

Using the final x position(120 meters) and the constant x velocity you found, you can find the time the ball be at that distance. Plug this time into the quation final vertical displacement= initial vertical displacemnt + (initial y-velocity x time) + (acceleration/gravity x time^2). if the answer is less than 3.8 meters, i didn't make it.

 

[The_big_dill]

Hi moondawg,
You solved this equation, and then doubled the time to find the time of 7s. But what exactly does this particular time represent? In other words, what is the ball doing (or what can you say about its position) at t=7s?

+1

If you can answer this question, you will (most likely) know what's going on. If you understand what is going on, it makes solving the problem a LOT easier.

 

[alphysicist]

Hi jmb88korean,

Using the final x position(120 meters) and the constant x velocity you found, you can find the time the ball be at that distance. Plug this time into the quation final vertical displacement= initial vertical displacemnt + (initial y-velocity x time) + (acceleration/gravity x time^2). if the answer is less than 3.8 meters, i didn't make it.

I believe that quation is missing a factor of 1/2.