Question

Romeo is throwing pebbles gently up to Juliet’s window, and he wants the pebbles to hit the window with only a horizontal component of velocity. He is standing at the edge of a rose garden 8.0 m below her window and 8.5 m from the base of the wall (Fig. 3-49). How fast are the pebbles going when they hit her window?

Short answer

The speed of the pebbles is $6.65\frac{\mathrm{m}}{\mathrm{s}}$when they hit the window.

Step-by-step solution

Step 1. Given data

The pebbles move forward due to the horizontal velocity component, and it remains the same throughout the motion. There is a downward gravitational acceleration.

Given data:

The height of the window is $y=8.0\mathrm{m}$.

The horizontal distance of Romeo from the base of the wall is $x=8.5\mathrm{m}$.

When the pebble hits the window, its vertical velocity is zero, i.e., the final vertical velocity of the pebbles is $v_y=0$.

Assumptions:

Let the pebbles take t time to hit the window. Let $v_{\mathrm{x}}$be the horizontal velocity of the pebbles throughout the motion. Also, let $v_{\mathrm{yo}}$be the initial vertical velocity of the pebbles.

Step 2. Analyze the horizontal motion

The time for the motion of the pebbles is $t=\frac{x}{v_{\mathrm{x}}}\dots \left(i\right)$

Step 3. Analyze the vertical motion

For the vertical displacement of the pebbles,

$\begin{array}{c}{v}_{\mathrm{y}}^2=v_{\mathrm{oy}}^2-2gy\\ 0^2=v_{\mathrm{oy}}^2-\left(2\times \left(9.80\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\times 8.0\mathrm{m}\right)\\ v_{\mathrm{oy}}=12.52\frac{\mathrm{m}}{\mathrm{s}}\end{array}$

Now, role="math" localid="1644918178037" $v_o=v_{\mathrm{oy}}-gt\dots \left(ii\right)$

Now, from equations (i) and (ii),

$\begin{array}{c}{v}_o=v_{\mathrm{oy}}-g\times \left(\frac{x}{v_{\mathrm{x}}}\right)\\ 0=12.52\frac{\mathrm{m}}{\mathrm{s}}-\left(9.80\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\times \left(\frac{8.5\mathrm{m}}{v_{\mathrm{x}}}\right)\\ \left(9.80\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\times \left(\frac{8.5\mathrm{m}}{v_{\mathrm{x}}}\right)=12.52\frac{\mathrm{m}}{\mathrm{s}}\\ v_{\mathrm{x}}=6.65\frac{\mathrm{m}}{\mathrm{s}}\end{array}$

Hence, the speed of the pebbles is $6.65\frac{\mathrm{m}}{\mathrm{s}}$when they hit the window.