# Simple question about parabolic trajectory

good night,

we are new in this forum.
we have a physics question about parabolic trajectory...
when a projectile is thrown with a certain angle, with initial v_0x and v_0y speeds, the trajectory of this projectile will be a parabola...
the initial x and y velocities will be:
$$v_{0x}=vcos\theta$$
$$v_{0y}=vsen\theta$$
and the x and y velocities after a certain time will be:
$$v_{x}=v_{0}t$$
$$v_{y}=v_{0y}-gt$$
is it right?
but here is our doubt:
we were wondering how we could calculate the work of this projectile.
could it be:
$$\tau=Fx=Fv_{0}t$$?
if our English is not very expressive, it's because we're Brazilian... we're not native English speakers.

The concept of work arises only when there is a force acting on a body.In case of a projectile no force acts in the X direction.But the force of gravity acts in the Y direction. So the work done may be F.y. It would be mg(component of displacement in the direction of force).I'm afraid you are wrong in mentionong that Vx=Vot.the horizontal component of velocity never changes.So atany time it would be Vocostheta

The concept of work arises only when there is a force acting on a body.In case of a projectile no force acts in the X direction.But the force of gravity acts in the Y direction. So the work done may be F.y. It would be mg(component of displacement in the direction of force).I'm afraid you are wrong in mentionong that Vx=Vot.the horizontal component of velocity never changes.So atany time it would be Vocostheta

thank you.
so, the work done by gravity against the projectile would be
$$W=Fy=mgy=mg(v_{y}t-\frac{1}{2}gt^{2})$$
right?
about the x velocity, we were very wrong in saying that. actually, $$v_{x}=v_{0}$$ at any time, as you said.
what we meant was:
$$x=v_{0}t=v_{x}t$$

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the work done by the projectile would be

Shouldn't that be phrased as "the work done by gravity against the projectile" ....?

Shouldn't that be phrased as "the work done by gravity against the projectile" ....?
thank you for pointing this mistake.
what should we do if we wanted to know the work done by the force applied to the ground by a projectile when it hits the ground (that is, in the maximum x distance)?

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Shooting Star
Homework Helper
Shouldn't that be phrased as "the work done by gravity against the projectile" ....?

It should be work done by gravity on the projectile.

thank you for pointing this mistake.
what should we do if we wanted to know the work done by the force applied to the ground by a projectile when it hits the ground (that is, in the maximum x distance)?

The energy of the projectile is dissipated once it comes to a stop after impact. This must be equal to the work done by the force the ground exerts on the projectile.

(This looks like a homework problem.)

It should be work done by gravity on the projectile.

The energy of the projectile is dissipated once it comes to a stop after impact. This must be equal to the work done by the force the ground exerts on the projectile.

(This looks like a homework problem.)

thank you for the response.
actually, no, it's not a homework problem, it is just a curiosity about a physics subject that we don't know completely but we would like to, that is why we are asking.
can you show it to us with mathematics?

just out of curiosity:
if someone throws an object in a certain angle, with a certain force, how can we measure its initial x and y velocities?
we imagine it could be like:
$$F=ma$$
if we know what force is applied, and the mass of the object, then we can discover its acceleration in the moment it is thrown. then, with the acceleration, we can discover the velocity (using Torricelli's equation).
knowing the initial velocity, and the angle of the throwing, we can measure the initial x and y velocities:
$$v_{0x}=v_{0}cos\theta$$
$$v_{0y}=v_{0}sen\theta$$
is that right?
if the object has a mass m and a force F is applied on it by the hand of the thrower at a certain angle, so can we discover the acceleration and then discover the initial velocity at which the object was thrown?

Shooting Star
Homework Helper
can you show it to us with mathematics?

I can just give you the very basics. If a body falls to the ground from a height h above the ground, then the initial potential energy mgh is converted into kinetic energy (½)mv^2, where v is the speed just before it hits the ground. Generally the body stops after travelling some distance into the ground, because of the resistive force exerted on the body by the ground. (I am not considering rebounds here.) Since energy is conserved, we know that the kinetic energy must have been dissipated into other forms of energy, mostly heat and sound, and deformation of the ground and/or the body. So, the amount of energy dissipated must be equal to the initial PE wrt the ground, which was mgh.

just out of curiosity:
if someone throws an object in a certain angle, with a certain force, how can we measure its initial x and y velocities?
we imagine it could be like:
$$F=ma$$
if we know what force is applied, and the mass of the object, then we can discover its acceleration in the moment it is thrown. then, with the acceleration, we can discover the velocity (using Torricelli's equation).
knowing the initial velocity, and the angle of the throwing, we can measure the initial x and y velocities:
$$v_{0x}=v_{0}cos\theta$$
$$v_{0y}=v_{0}sen\theta$$
is that right?
if the object has a mass m and a force F is applied on it by the hand of the thrower at a certain angle, so can we discover the acceleration and then discover the initial velocity at which the object was thrown?
Regarding the last question that you’ve asked, if the hand is assumed to be imparting a constant force to the ball, then it would be possible to calculate the initial velocity in the way you have said. (I had to look up what Torricelli’s equation was! )