Kinematics of a Home Run: What is the Initial Speed and Components at the Fence?

[Jabababa]

Homework Statement

a baseball player hits a home run that just clears a fence that is 19.0 meters high.The fence is located 128 meters from the home player(where the ball is hit). The ball was hit at an angle of 37.0 degrees above the horizontal from a height of 0.900 meters above the ground.

a) what is the initial speed of the ball?
b)how long does it take the baseball to reach the fence
c)what are the horizontal and verticle component of the ball and its speed when it reaches the fence?

Homework Equations

The Attempt at a Solution

for a), it confuses me because of the 0.900m it was hit above the ground. Is it possible to make the 0.900 into 0 and the fence is 19.0m - 0.900m? So everything is shifted to ground level.

 

[PeterO]

Homework Statement

a baseball player hits a home run that just clears a fence that is 19.0 meters high.The fence is located 128 meters from the home player(where the ball is hit). The ball was hit at an angle of 37.0 degrees above the horizontal from a height of 0.900 meters above the ground.

a) what is the initial speed of the ball?
b)how long does it take the baseball to reach the fence
c)what are the horizontal and verticle component of the ball and its speed when it reaches the fence?

Homework Equations

The Attempt at a Solution

for a), it confuses me because of the 0.900m it was hit above the ground. Is it possible to make the 0.900 into 0 and the fence is 19.0m - 0.900m? So everything is shifted to ground level.

I would certainly approach it that way. But I would just consider zero altitude to be 0.900m above the ground rather than actually moving to ground level.

 

[Jabababa]

I would certainly approach it that way. But I would just consider zero altitude to be 0.900m above the ground rather than actually moving to ground level.

Yes, that's what i mean heh. Not exactly moving them and changing the question.

For part b it is easy if i have the initial velocity and the angle, i can just use the horizontal equation ( x= (Vo)(T) to find the time.

For part c, I know how to find the final velocity, but how do i find the vertical and horizontal component if i don't know the angle of the ball at that point?

 

[PeterO]

Yes, that's what i mean heh. Not exactly moving them and changing the question.

For part b it is easy if i have the initial velocity and the angle, i can just use the horizontal equation ( x= (Vo)(T) to find the time.

That would make it easy - however you don't have the initial velocity - that's what part a is all about

For part c, I know how to find the final velocity, but how do i find the vertical and horizontal component if i don't know the angle of the ball at that point?

I think you are heading for some simultaneous equations.
Vertically, the ball will go to maximum height then back to fence height in a certain time.
In that same time it will cover the 128m horizontally
I suspect you will have to put those two ideas together in some simultaneous equations.

Note: The 37 degree angle is very close to one of the angles in a 3-4-5 right angled triangle. Not sure whether you are allowed/expected to approximate to that. (it is actually 36 degrees 52 minutes)

 

[Jabababa]

I think you are heading for some simultaneous equations.
Vertically, the ball will go to maximum height then back to fence height in a certain time.
In that same time it will cover the 128m horizontally
I suspect you will have to put those two ideas together in some simultaneous equations.

Note: The 37 degree angle is very close to one of the angles in a 3-4-5 right angled triangle. Not sure whether you are allowed/expected to approximate to that. (it is actually 36 degrees 52 minutes)

well no, for part a) i just find the initial velocity...and then use that initial velocity to calculate part b). We have x,initial velocity(from part a) and the angle so its just x = Vocos*t -> t= x/VoCos. Isnt that how you do it for horizontal, since horizontal velocity is constant.

 

[PeterO]

well no, for part a) i just find the initial velocity...and then use that initial velocity to calculate part b). We have x,initial velocity(from part a) and the angle so its just x = Vocos*t -> t= x/VoCos. Isnt that how you do it for horizontal, since horizontal velocity is constant.

If you can find the initial velocity then go ahead. I am just not sure how you proposed to do that.

If the initial velocity is 50 m/s, then the horizontal component is approx 40 m/s, so it takes just over 3 seconds to reach the fence.

If the initial velocity is 25 m/s, it will take just over 6 seconds to reach the fence.

If the initial velocity is 50 m/s, then the vertical velocity is approx 30 m/s.
In just over 3 seconds (see above) the ball will be very close to maximum height - which is approx 45 m - so the ball will sail high over the fence.

If the initial velocity is 25 m/s, the initial vetical velocity is only approx 15 m/s, adn when considering the 6 seconds (mentioned above) the ball will have hit the ground - short of the fence.

You need to calculate the balancing act so that the ball just clears the fence.

That is where I think the simultaneous equations will come in.

 

[Jabababa]

alright I just got the velocity and time. Can anyone explain part c to me please. I know how to find the final velocity with the height, initial velocity, the angle and the time, but how do I find the x and y component if I don't know the angle when it just got pass the fence. Thank you.

 

[PeterO]

alright I just got the velocity and time. Can anyone explain part c to me please. I know how to find the final velocity with the height, initial velocity, the angle and the time, but how do I find the x and y component if I don't know the angle when it just got pass the fence. Thank you.

The horizontal component is the same it has always been.
The vertical component is equal to a similar body, projected straight up with the initial vertical component, which has reached a height of (19-0.900) metres - whether on the way up oR on the way down (the magnitude is the same, only the direction varies).