A large horizontal platform, open at the top, moves vertically upward with a constant velocity $v$. A small object is launched from the platform at an angle of $\theta = 30^\circ$ with a launch velocity $u$ (measured relative to the platform). Determine the height above the launch point at which the object strikes the platform. (Ignore air resistance.)
Solution
We can solve this problem by analyzing the motion in the stationary ground reference frame as a function of time $t$.
Step 1: Position of the Platform
Since the platform moves upward with a constant velocity $v$, its vertical position $z_p$ at any time $t$ is given by:
$$z_p = vt$$Step 2: Position of the Object
The initial vertical velocity of the object in the ground frame is the sum of the platform's velocity and the vertical component of the object's launch velocity: $u \sin\theta + v$.
Accounting for the acceleration due to gravity $g$, the vertical position $z_o$ of the object at time $t$ is:
$$z_o = (u \sin\theta + v)t - \frac{1}{2}gt^2$$Step 3: Determining the Time of Impact
The object strikes the platform when their vertical positions are equal ($z_p = z_o$):
$$vt = (u \sin\theta + v)t - \frac{1}{2}gt^2$$Subtracting $vt$ from both sides simplifies the equation to:
$$u \sin\theta \, t = \frac{1}{2}gt^2$$For the non-zero time of impact ($t \neq 0$), we can solve for $t$:
$$t = \frac{2u \sin\theta}{g}$$Step 4: Calculating the Height
Substitute the time of flight $t$ back into the equation for the platform's height:
$$z = vt = v \left(\frac{2u \sin\theta}{g}\right) = \frac{2uv \sin\theta}{g}$$Substituting the given angle $\theta = 30^\circ$ (where $\sin 30^\circ = \frac{1}{2}$):
$$z = \frac{2uv \left(\frac{1}{2}\right)}{g} = \frac{uv}{g}$$Final Answer
The object will strike the platform at a height of $\frac{uv}{g}$ above the launch point.