Simple Projectile Motion Problem

In summary, the ball is hit at 60.0m/s at an angle of 30 degrees to the horizontal. The time ball was in the air and how long it was in the air, how far it went, and the ball's vertical displacement were all calculated.
  • #1
PinguNinja
5
0

Homework Statement


Baseball is hit at 60.0m/s at an angle of 30 degrees to the horizontal. Assuming the field is level find:
Time ball was in air
How long it was in the air
How far did it go
Given
Assume up is +ve
Calculated data within brackets
Horizontal
v=(52m/s)
d=
t=
Vertical
vi=(30m/s)
vf=
a=-9.8m/s
d=
t=
Ball
v=60.0m/s[Forward30Up]

Homework Equations


sinθ=opp/hyp
cosθ=adj/hyp
d=ut+0.5at^2
v=d/t

The Attempt at a Solution


Assume (variable)(h) means a horizontal variable and (variable)(v) means the vertical variable.
So i drew out the vector diagram for the ball, with v(ball)=v(h)+v(v), knowing angle is 30
Using sine and cosine i was able to determine the values for v(h)=52m/s and v(v)=30m/s.
(Calculated info written in the givens within brackets)
So now I am stuck...I am not given enough information to determine any other variables, I think?
 
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  • #2
Welcome to PF!

Hi PinguNinja! Welcome to PF! :wink:
PinguNinja said:
Horizontal
v=(52m/s)
d=
t=
Vertical
vi=(30m/s)
vf=
a=-9.8m/s
d=
t=

Yes, that's fine so far. :smile:

Now use the standard constant acceleration equation in the vertical direction, to find t.

(the displacement d = 0 :wink:)​
 
  • #3
Perhaps this is a silly question, but why would the displacement be zero? Were the displacement be zero, would that not mean that the ball is at the start position? (Start position being where the ball is hit). If we were to set the point it hit the ground to be zero, then would we not run into the problem where we do not know at what height the start position should be?

d=ut+0.5at^2
0=30t+0.5(-9.8)t^2
-30t=0.5(-9.8)t^2
6.1=t
 
  • #4
Assuming the ball is initially on the ground (call this h = 0) and it is hit with a bat at an angle of 30 degrees above the horizontal we can say that the displacement is 0 becasue the ball travels up in the air to its max height (call this h = [itex]h_{max}[/itex]; it then falls to the ground returning to h = 0 - so the displacement of the ball is 0 (it returned to the height from which it was hit).

If you were to hold the ball above the ground at some arbitrary height (say, h = [itex]h_{o}[/itex] it would rise to its max height [itex]h_{max}[/itex], then fall to its original height [itex]h_{o}[/itex] and continue falling until it hit the ground. In this case the total displacement would be [itex] - h_{o}[/itex] because the change in its position would be [itex]\Delta h = h_{final} - h_{initial} = 0 - h_{o}[/itex].
 
  • #5
Alright. So since it appears that the question is not doable without assumptions (i assume? :wink:) Then I will assume that the ball hits the bat when the bat is touching the ground :3.
 
  • #6
That seems like the best assumption to make, in my opinion. :)
 
  • #7
Hi PinguNinja! :smile:

(just got up :zzz:)

Yes, I agree with Tsunoyukami :smile: … the question is badly worded. :frown:

However, you can work it out from …
PinguNinja said:
Assuming the field is level …

clearly this was put there for some purpose,

and the writer obviously thought he was saying that the vertical displacement was zero! :biggrin:
 

Related to Simple Projectile Motion Problem

What is Simple Projectile Motion?

Simple Projectile Motion is the motion of an object that is thrown or projected into the air at an angle, and then moves along a curved path under the influence of gravity.

What is the formula for calculating Simple Projectile Motion?

The formula for calculating Simple Projectile Motion is: - Horizontal displacement: x = v0 * cos(theta) * t- Vertical displacement: y = v0 * sin(theta) * t - 0.5 * g * t^2- Time of flight: t = 2 * v0 * sin(theta) / g- Maximum height: hmax = v0^2 * sin^2(theta) / 2g

What is the difference between initial velocity and final velocity in Simple Projectile Motion?

Initial velocity (v0) is the velocity at which an object is launched or thrown, while final velocity is the velocity of the object at a specific point in time during its trajectory. In Simple Projectile Motion, final velocity is the vector sum of the horizontal and vertical components of the initial velocity.

What are some real-life applications of Simple Projectile Motion?

Simple Projectile Motion has many real-life applications, including:- Ballistics: determining the trajectory of projectiles such as bullets or missiles.- Sports: understanding the motion of objects such as balls in sports like basketball, soccer, or baseball.- Fireworks: predicting the path and height of fireworks during a display.- Motion of planets and satellites: predicting the motion of objects in space, such as planets and satellites.

What factors can affect Simple Projectile Motion?

Some factors that can affect Simple Projectile Motion include: - Initial velocity: the speed and direction at which the object is launched.- Angle of projection: the angle at which the object is launched.- Air resistance: the force that opposes the motion of the object through the air.- Gravity: the force that pulls the object towards the Earth.- Mass and size of the object: heavier and larger objects may have different trajectories compared to lighter and smaller ones.

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