Calculating Trajectory & Time of Flight

In summary, the conversation discusses how to approach a physics problem involving a golf ball being launched from a hill at a certain angle and initial velocity. The conversation delves into the use of trajectory equations and finding the equation for the parabolic arc of the ball's movement. The final solution involves finding the height of the hill, which is found to be 107 meters.
  • #1
XwakeriderX
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0

Homework Statement



See picture

Homework Equations


Trajectory equation and total time of flight



The Attempt at a Solution


Okay what would be the best way to approach this? should i find out the distance the ball goes? then subtract 35m to to get my length of hill and use that for cos30=x/h
 

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  • #2
Please anyone I've been at this for hours
 
  • #3
Which trajectory equations do you think apply? Note: I see from your prior posts that the problems you are submitting are of a somewhat higher level of difficulty than introductory level basic problems. What level of physics are you taking?
 
  • #4
Just a normal college physics class "Mechanics-Solids/Fluids" at a junior college. My teacher does not give any examples or anything. This is his old test questions so I am trying to figure them out because they are not like normal two dimension or trajectory problems! Are you not watching the game!? i have to study for physics my test is on wednesday! Okay now back on topic

I know the basic trajectory equations

Y=Yo + Xtan(THETAo)- (gx^2/(2Vo^2Cos^2THETAo)

I think i want to see where the golf ball is at the base of the hill an then add the rising mountain to the downward acceleration
 
  • #5
I would be watching the game, but I'm trying to assist you! But in the meantime , I have to run some errands, so rather than leave you hanging, I'll see if someone else can assist. Hang in there.
 
  • #6
thanks! i really appreciate it! I'm going premed but UCLA wants me to have physics under by belt which doesn't really make sense unless i want to find out how far a kid fell of his bike to break his arm...haha
 
  • #7
XwakeriderX said:

Homework Statement



See picture

Homework Equations


Trajectory equation and total time of flight



The Attempt at a Solution


Okay what would be the best way to approach this? should i find out the distance the ball goes? then subtract 35m to to get my length of hill and use that for cos30=x/h

What game?

One key hint on trajectory questions like this is to see that the horizontal velocity remains constant when there is no air resistance. The vertical velocity does vary, according to the constant acceleration kinematic equations. So the ball follows a parabolic arc. You need to find the equation for that arc, and equate it with the equation of the line for the hill's slope. Where the two lines meet, that's where the ball hits.

See "Kinematics of constant acceleration" on this page:

http://en.wikipedia.org/wiki/Kinematics

.
 
  • #8
red sox!

Okay that makes some sense to me, ill try finding the equation to the arc
 
  • #9
Range = [sin(2theta)V(initial)^2]/ acceleration ?
 
  • #10
Set up your coordinates so the point (x,y)=(0,0) is at the base of the hill. The hill is then given by the line y=mx for some value of m, which you can figure out. Tell us what the equations for x(t) and y(t) of the ball should be.
 
  • #11
d = 35.0 m,
angle A = 45.0 degrees,
angle B = 30.0 degrees,
initial velocity v = 145*0.447 m/s
find h

horizontal displacement x = d + h/tan(B) = vcos(A)*t (1)
vertical displacement y = h = vsin(A)*t - gt^2/2 (2)
from (2),
gt^2/2 - vsin(A)*t + h = 0
solve it and get
t = {vsin(A) + sqrt[v^2*sin(A)^2 - 2gh]}/g
put it into (1), get
d + h/tan(B) = vcos(A)*{vsin(A) + sqrt[v^2*sin(A)^2 - 2gh]}/g
gd + gh/tan(B) = v^2*cos(A)*sin(A) + v*cos(A)*sqrt[v^2*sin(A)^2 - 2gh]}
insert known values, simplify and get
sqrt(2100 - 19.6h) = 0.37037h - 38.337
squared both sides and simplify,
0.13717h^2 - 8.79h - 630.3 = 0
solve it and get
h = 107 m

:)
 

FAQ: Calculating Trajectory & Time of Flight

1. How do you calculate trajectory and time of flight?

Trajectory and time of flight can be calculated using the equations of motion, specifically the projectile motion equations. These equations take into account the initial velocity, angle of launch, and acceleration due to gravity to determine the trajectory and time of flight for a projectile.

2. What factors affect the trajectory and time of flight of a projectile?

The trajectory and time of flight of a projectile are affected by several factors, including the initial velocity of the projectile, the angle of launch, and the presence of air resistance. Other factors such as wind speed and direction can also impact the trajectory and time of flight.

3. How does air resistance affect the trajectory and time of flight?

Air resistance, also known as drag, can have a significant impact on the trajectory and time of flight of a projectile. As the projectile travels through the air, it experiences a force in the opposite direction of its motion, which can cause it to slow down and deviate from its intended trajectory. This results in a shorter time of flight and a different trajectory than if there were no air resistance.

4. Can you calculate trajectory and time of flight for a projectile on other planets?

Yes, the same equations used to calculate trajectory and time of flight on Earth can also be applied to other planets. However, the values for acceleration due to gravity and air resistance may be different on other planets, so these factors must be taken into account when calculating trajectory and time of flight.

5. How can trajectory and time of flight calculations be used in real-world applications?

Trajectory and time of flight calculations are used in many real-world applications, such as ballistics and aerospace engineering. They are also important in sports, such as baseball and golf, where understanding the trajectory and time of flight of a projectile can help players make more accurate shots. Additionally, these calculations are used in the design and testing of rockets, missiles, and other projectiles.

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