Velocity problem determining x and y components

[brandon133]

An object undergoing parabolic motion travels 100 m in the horizontal direction before returning to its initial height. If the object is thrown initially at a 30° angle, determine the x component and the y component of the initial velocity. (Neglect any effects due to air resistance.)

I am not sure how to go about solving... I got an initial vector from the throw but I don't think it can be the initial velocity vector since I have no judgment of time.. could someone show me the process of solving this?

 

[HallsofIvy]

If we let "$v_x$" and "$v_y$" be the components of velocity in the x and y directions, respectively, then $x= v_xt$ and $y= -4.9t^2+ v_yt$ so we have $y= -4.9t^2+ v_yt= 0$ (the projectile hits the ground) and $x= v_xt= 100$ (the projectile hits the groud after 100 m). The fact that "the object is thrown initially at a 30° angle" means that $v_y/v_x= tan(30)$. That gives you three equations to solve for the three values, t, $v_x$, and $v_y$.

 

[rl.bhat]

If vsinθ is the y component of the velocity, what is the time the projectile takes to return to its Initial height?
If vcosθ is the x component of the velocity, what is the expression for the range x?

 

[azizlwl]

An object undergoing parabolic motion travels 100 m in the horizontal direction before returning to its initial height. If the object is thrown initially at a 30° angle, determine the x component and the y component of the initial velocity. (Neglect any effects due to air resistance.)

I am not sure how to go about solving... I got an initial vector from the throw but I don't think it can be the initial velocity vector since I have no judgment of time.. could someone show me the process of solving this?

The time taken for the object to travel horizontally 10m is equal to the time taken from launching up and back to same level.

It is a single body, at same place(different x and y coordinates) and at same time.

 

[Nickie]

Sure, I'd be happy to help with this problem! To solve for the x and y components of the initial velocity, we can use the equations for projectile motion. Let's start by defining some variables:

x0 = initial horizontal position (0 m)
y0 = initial vertical position (0 m)
x = final horizontal position (100 m)
y = final vertical position (0 m)
θ = initial angle of the throw (30°)
v0 = initial velocity (unknown)
g = acceleration due to gravity (-9.8 m/s^2)

Now, we can use the equations of motion to solve for the initial velocity. The first equation we will use is the horizontal position equation:

x = x0 + v0x * t

Since the initial horizontal position is 0 m, this equation simplifies to:

x = v0x * t

We can also use the vertical position equation:

y = y0 + v0y * t + 1/2 * g * t^2

Since the initial vertical position is also 0 m, this equation simplifies to:

y = v0y * t + 1/2 * g * t^2

We can use the fact that the object returns to its initial height to set y equal to 0. This allows us to solve for the time it takes for the object to travel 100 m horizontally:

100 = v0x * t

t = 100 / v0x

Now, we can plug this value for t into the vertical position equation and solve for the initial vertical velocity:

0 = v0y * (100 / v0x) + 1/2 * g * (100 / v0x)^2

0 = v0y + 1/2 * g * (100 / v0x)

v0y = -1/2 * g * (100 / v0x)

Finally, we can use trigonometry to relate the initial velocity to its x and y components:

v0x = v0 * cos(θ)
v0y = v0 * sin(θ)

Substituting these values into the equation for v0y, we get:

v0 * sin(θ) = -1/2 * g * (100 / v0 * cos(θ))

Solving for v0, we get:

v0 = -100 * tan(θ) / g

Now,