<li>*As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length of your window is L_w. Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity g.</li>
</ol>
<p>Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward).*</p>
<p>From what height h above the bottom of your window was the flower pot dropped? Express the answer in terms of L_w, t, and g.</p>
<li>Two basketball players are essentially equal in all respects. (They are the same height, they jump with the same initial velocity, etc.) In particular, by jumping they can raise their centers of mass the same vertical distance, H (called their “vertical leap”). The first player, Arabella, wishes to shoot over the second player, Boris, and for this she needs to be as high above Boris as possible. Arabella jumps at time t=0, and Boris jumps later, at time t_R (his reaction time). Assume that Arabella has not yet reached her maximum height when Boris jumps.</li>
</ol>
<p>Find the vertical displacement D(t) = h<em>A(t) - h</em>B(t), as a function of time for the interval 0 < t < t<em>R, where h</em>A(t) is the height of the raised hands of Arabella, while h_B(t) is the height of the raised hands of Boris.</p>
<ol>
<li><p>Use s = ut + (0.5)(a)t^2 where s=distance, u=init.velocity, a=accel</p>
<p>The pot travels a distance of L<em>w from the top ->bottom of your window, and takes t seconds to do it; so
L</em>w = ut + (0.5)gt^2, from which
u = (L_w - 0.5gt^2)/t</p>
<p>The pot was travelling at a velocity of u at the top of your window, and started at zero velocity when it was (h - Lw) length units from the <em>top</em> of your window. Suppose it took y time units to reach the top of your window. Now use<br>
final<em>velocity = init</em>velocity + (accel)(time)
or u = 0 + gy
or y = u/g</p></li>
</ol>
<p>Now use the distance formula
h - L<em>w = (0)(y) + (0.5)(g)y^2
or h = L</em>w + (0.5)(g)(u/g)^2
= L<em>w + ((0.5)/g)* ((L</em>w - 0.5gt^2)/t)^2</p>
<ol>
<li>Since both players are the same height, the 'vertical displacement' defined is the same as the diff. in height between their shoe soles. </li>
</ol>
<p>If 0 < t < t<em>R, Boris is stationary, and D(t) is simply the height of Alice's jump as a function of time i.e. D(t) = h</em>A(t) - 0 = ut - 0.5gt^2 .
(Note that acceleration here is -g, since we are measuring distance upwards as +ve, and gravity acts in the opposite direction).</p>
<p>[For a harder version of the problem, try to find D(t) when t > t_R.
Then,</p>
<p>h<em>A(t) = ut - (0.5)gt^2<br>
h</em>B(t) = u(t - t<em>R) - (0.5)(g)(t-t</em>R)^2
= ut - ut<em>R - (0.5)(g)(t^2 - 2tt</em>R + t<em>R^2)
so D(t) = h</em>A(t) - h<em>B(t)
= ut - (0.5)gt^2 - (ut - ut</em>R - (0.5)(g)(t^2 - 2tt<em>R + t</em>R^2))
= ut - (0.5)gt^2 - ut + ut<em>R + (0.5)(g)(t^2 - 2tt</em>R + t<em>R^2)
= ut</em>R + (0.5)(g)( - 2tt<em>R + t</em>R^2)) ]</p>
<p>I got the first part of this question, but I'm having problems with the second part.</p>
<ol>
<li>A cannon, located 60.0 m from the base of a vertical 25.0m-tall cliff, shoots a 15-kg shell at 43.0 degrees above the horizontal toward the cliff.</li>
</ol>
<p>The first part asks "What must the minimum muzzle velocity be for the shell to clear the top of the cliff?" I got 32.6 m/s, which is correct.</p>
<p>However, the second part asks "The ground at the top of the cliff is level, with a constant elevation of 25.0 m above the cannon. Under the conditions of part (a), how far does the shell land past the edge of the cliff?"</p>
<p>I got 49.2 m twice, but that is not correct.</p>
<p>What's the right answer and how do I do it? I tried breaking up the minimum velocity into x and y components and figuring out the time required for the shell to clear the cliff using the equation (-1/2)(9.80(t ^2) + (v,y)(t) -25.0 = 0. I then multiplied this time by the x component of the minimum velocity and subtracted 60 from my answer.</p>
<p>You first need to compute t1 (=time to reach the lip of the cliff) and t2 (=time to reach max. elevation). The required answer is 2 (x_component of velocity) (t2-t1), since the time to go from 'touching' the lip of the cliff to max.height is the same as the time to fall back from max.height to the ground at cliff level. </p>
<p>You can get t1 as 60/ (x<em>component of muzzle velocity). You can get t2 by solving v = u + at or 0 = (y</em>component of muzzle velocity) - g.t . See if this works.</p>