Projectile Motion Unusual Problem

[Georgiana.B]

Homework Statement

A body which is thrown upward from the ground at an angle θ passes through 2 points situated at height h at times [t][1] and [t][2] from launching.Calculate the initial velocity [v][0] and the height h.

Homework Equations

The equation of motion for a projectile motion:
x(t)= [x][0]+ [V][0](t-[t][0])- g/2[(t-[t][0])][2]
V=[V][x] * cosθ +[V][y] * sinθ

The Attempt at a Solution

I thought that since the body passes through the 2 points which are situated at the same height at time [t][1] and [t][2], at means that y([t][1])=y ([t][2]). So, by replacing it in the formula above mentioned with [V][y] * sinθ as [V][0], i tried to get [V][0] with respect to time,but that doesn't seem to help since i have no numbers to relay on.Could anyone please share any ideas of solving?

 

[gneill]

Homework Statement

A body which is thrown upward from the ground at an angle θ passes through 2 points situated at height h at times [t][1] and [t][2] from launching.Calculate the initial velocity [v][0] and the height h.

Homework Equations

The equation of motion for a projectile motion:
x(t)= [x][0]+ [V][0](t-[t][0])- g/2[(t-[t][0])][2]
V=[V][x] * cosθ +[V][y] * sinθ

The Attempt at a Solution

I thought that since the body passes through the 2 points which are situated at the same height at time [t][1] and [t][2], at means that y([t][1])=y ([t][2]). So, by replacing it in the formula above mentioned with [V][y] * sinθ as [V][0], i tried to get a express [V][0] with respect to time,but that doesn't seem to help since i have no numbers to relay on.Could anyone please share any ideas of solving?

Hi Georgiana.B, Welcome to Physics Forums.

If you have the general equation of motion for the y-direction and two solution pairs [h,t1] and [h,t2], then you can write two equations involving all the given variables. See if you can't then solve for Vo in terms of the given variables. You should be able to eliminate h from the Vo solution, too

Note that your solution will involve variables t1, t2, θ, and constant g.

 

[haruspex]

y([t][1])=y ([t][2]). So, by replacing it in the formula above mentioned with [V][y] * sinθ as [V][0], i tried to get a express [V][0] with respect to time,but that doesn't seem to help since i have no numbers to relay on.

You know more than merely the two heights are the same. You know that height equals h. Your answer will be in terms of h.

 

[Georgiana.B]

A better question would be,which are other ways of solving this kind of problems?

 

[haruspex]

A better question would be,which are other ways of solving this kind of problems?

Depends what you mean by 'this kind of problem'. If you mean constant acceleration problems then the SUVAT equations, such as you quoted, are the natural choice. The trick, often is to pick the most useful for the question. Of the usual five variables - distance, acceleration, time, start velocity, end velocity - each equation references four. Typically you know three and want a fourth, which tells you which equation to use.
In some problems there can be a quicker way, perhaps using symmetry.
In the present problem, you could start by finding the vertical launch speed needed to keep an object in the air for time t₂-t₁. From that, you can easily find the vertical speed needed when launched from h lower down.