Projectile motion of a ball

  • Thread starter ezsmith
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  • #1
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Homework Statement



This is a review question from the test I did few months back, and I need to review it for my finals.

A ball is thrown from the top of a building 35m high with an initial speed of Vi at an angle 30° above the horizontal. The ball strikes the ground at a point 80m from the base of the building. Find

a) the time of the ball is in the flight
b) the horizontal and vertical component of the initial velocity
c) its initial velocity and
d) the velocity of the ball just before it strikes the ground


Homework Equations



Sy= Viy t + 1/2Ayt^2

The Attempt at a Solution



My attempt was resolving into x and y direction, so,

Ax= 0 Ay=-9.8m/s
Sx= 80m Sy=-35, Viy= 0

Therefore: Sy= Viy+1/2Ayt^2
-35= 0 + 1/2(-9.8)t^2
Finally: My t = 2.67s

Apparently, I was marked wrong, so my time I found is wrong and therefore I couldn't proceed with b,c and d with the wrong time.

Help would be really appreciated! Thanks.
 

Answers and Replies

  • #2
SteamKing
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The problem clearly states that the ball is thrown at initial speed Vi at an angle of 30 degrees above the horizontal.

Why did you take Viy = 0? When is Viy = 0 during the flight of the ball?
 
  • #3
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The problem clearly states that the ball is thrown at initial speed Vi at an angle of 30 degrees above the horizontal.

Why did you take Viy = 0? When is Viy = 0 during the flight of the ball?

Oh, I think I figured it out

So its, I use Vx = Vicos30°
V = s/t
V = 80 / t cos30°

Then with this equation V = 80 / t Vicos30° I sub into my component y which is
-35 = Vi sin30 t-4.9t^2
-35 = 80 sin30 t/ t cos30° - 4.9t^2
t = 4.07s

Is it right?
 
  • #4
SteamKing
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I don't know why you are using V = s/t here. The initial velocity Vi is a given.

The path of the ball is analyzed in two parts: the first part is the ball travelling upward until gravity reduces the vertical velocity to zero. The second part starts at this point as the ball free falls from the apex of its trajectory and hits the ground.

A simple sketch would be most helpful in analyzing this type of problem, rather than just throwing a bunch of random formulas at it.
 
  • #5
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I don't know why you are using V = s/t here. The initial velocity Vi is a given.

The path of the ball is analyzed in two parts: the first part is the ball travelling upward until gravity reduces the vertical velocity to zero. The second part starts at this point as the ball free falls from the apex of its trajectory and hits the ground.

A simple sketch would be most helpful in analyzing this type of problem, rather than just throwing a bunch of random formulas at it.

For my vertical component, its -35 = Vi sin30 t -4.9t^2
At this stage, how am I going to solve with Vi being an unknown instead of a value?
 
  • #6
SteamKing
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For my vertical component, its -35 = Vi sin30 t -4.9t^2
At this stage, how am I going to solve with Vi being an unknown instead of a value?

Vertical component of what? The 35 m building is just sitting there.

You'll have to solve for time in terms of Vi, since a numerical value is not disclosed.
 
  • #7
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Vertical component of what? The 35 m building is just sitting there.

You'll have to solve for time in terms of Vi, since a numerical value is not disclosed.

After thinking real hard for almost 2 days.

I know that my horizontal and vertical component has the same time is in flight. So I can use resolve my t initially in the horizontal component of the flight using Sx = Vicos30 t right. So wouldn't be
t = Sx / Vi cos 30

Knowing my Sx is 80 and cos 30 is 0.87
Hence t = 92.38 / Vi
The problem is I can't eliminate my Vi. So like you said.. I treat my time in terms of Vi. Therefore my answer is t = 92.38 / Vi ??
 

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