# Projectile motion

Watch
Announcements
#1
At a point O on its path, a projectile has speed V and is travelling at an angle a above the horizontal. Derive the equation of the trajectory of the projectile in the form

y = x*tan(a) - gx2*(1+tan2(a))/(2V2)

Now, an aircraft is flown on the path given by y = 0.28x - (2.35*10-4) x2. By comparing the equation with the standard trajectory equation found above, find the speed and direction of motion of the aircraft at the point (0,0). Calculate the time of flight between the two points on the aircraft's path where y = 0.

If the equation is given numerically, doesn't this mean tan(a)=0.28 constantly? Which would mean that the motion is not projectile at all since the direction is constant ... what's going on here? How should I try to solve the problem?

As for the first part, I don't have a clue how to get there. Maybe there is a way to solve it with energy conservation? In my original attempt, I noted V*sin(a) = u*sin(theta) - gt where theta is the initial angle of projection and then tried to substitute into y = u*sin(theta)*t - 1/2gt2, where t was replaced in terms of x, but somehow ended up with

y = x*tan(a) + gx2*(1+tan2(a))/(2V2)

i.e. the wrong sign on the x2 term!
0
6 years ago
#2
At a point O on its path, a projectile has speed V and is travelling at an angle a above the horizontal. Derive the equation of the trajectory of the projectile in the form

y = x*tan(a) - gx2*(1+tan2(a))/(2V2)

Now, an aircraft is flown on the path given by y = 0.28x - (2.35*10-4) x2. By comparing the equation with the standard trajectory equation found above, find the speed and direction of motion of the aircraft at the point (0,0). Calculate the time of flight between the two points on the aircraft's path where y = 0.

If the equation is given numerically, doesn't this mean tan(a)=0.28 constantly? Which would mean that the motion is not projectile at all since the direction is constant ... what's going on here? How should I try to solve the problem?

As for the first part, I don't have a clue how to get there. Maybe there is a way to solve it with energy conservation? In my original attempt, I noted V*sin(a) = u*sin(theta) - gt where theta is the initial angle of projection and then tried to substitute into y = u*sin(theta)*t - 1/2gt2, where t was replaced in terms of x, but somehow ended up with

y = x*tan(a) + gx2*(1+tan2(a))/(2V2)

i.e. the wrong sign on the x2 term! so subbing in we have  How did you manage to get the wrong sign?
0
#3
You've given the expression in terms of θ, which appears to the angle of projection (since you used Vsin(θ)-gt as velocity in the y-direction) but I think the question asks for angle a instead, i.e. x,y as a function of the angle of motion of the projectile at that particular moment in time, rather than of initial projection?
0
6 years ago
#4
You've given the expression in terms of θ, which appears to the angle of projection (since you used Vsin(θ)-gt as velocity in the y-direction) but I think the question asks for angle a instead, i.e. x,y as a function of the angle of motion of the projectile at that particular moment in time, rather than of initial projection?
Then just replace theta with alpha. The working is the same we are just using a different staring point.
0
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (143)
14.3%
I'm not sure (42)
4.2%
No, I'm going to stick it out for now (301)
30.1%
I have already dropped out (26)
2.6%
I'm not a current university student (488)
48.8%