Question

Two tourists are standing at the top of the Burj Khalifa (the tallest building in the world). This mammoth spire stretches a staggering 828 meters into the sky, and these two are about to (very illegally) throw rocks off the edge and watch them fal Soon-to-be-arrested tourist A drops the rock. Soon-to-be-arrested tourist B decic to throw the rock downward at 20 degrees below the horizontal at 15 m/s. Your job: Calculate the difference in the time between the two rocks' impacts into the pavement below. Ignore air resistance for now. That'll come next exam. :) Fun fact: The result is kind of surprising. You can see how much initial velocity matters when distances are larger scaled like this. 0.513 s 0.041 s 12.999 s 12.495 s

Answer 1

First, let's calculate the time taken for the two rocks to drop to the ground.

Tourist A

Tourist A drops the ball vertically down with no initial velocity. We can find the time taken by the ball to reach the pavement can be calculated using the equation

$S=ut+\frac12 at^2$

Here, we have

$S=828\textrm{ m}$

$u=0\textrm{ m/s}$

$a=9.81\textrm{ m/s}^2$

So,

9.81 828 =0+一 七 2

$t_A=\sqrt{\frac{2\times828}{9.81}}=12.9926\textrm{ s}$

Tourist B

Tourist B drops throw the rock downward at 20 degrees below the horizontal at 15m/s. We need to calculate the time taken by the rock to drop, so we are only interested in the vertical component of this initial velocity.

The vertical component is

$u=15sin20=5.1303\textrm{ m/s}$

So, we have

$S=828\textrm{ m}$

$u=5.1303\textrm{ m/s}$

$a=9.81\textrm{ m/s}^2$

So,

$828=5.1303t+\frac{9.81}2t^2$

Using quadratic formula

$t_B=\frac{-5.1303+\sqrt{5.1303^2+16245.36}}{9.81}=12.4801\textrm{ s}$

So, the difference is

$\Delta t=t_A-t_B=12.9926-12.4801=0.5125\textrm{ s}\approx 0.513\textrm{ s}$

So, the correct answer is 0.513 s.

Thank you for posting your problem to Chegg.
If you liked my solution to your problem, a thumbs up will be highly appreciated. It motivates me to do more!