Question

Bill and Ted are standing on a bridge 40 ft above a river. Bill drops a stone, while Ted decides to throw a stone downward at 10 m/s. How long after Bill released his rock should Ted throw his if they want the stones to hit the water simultaneously? A. del t 15.5 sec B. del t 0.86 sec C. Cannot be determined. D. del t - 0.72 sec

Answer 1

Answer:

D.

Explanation:

In order to know how long after Bill released his rock should Ted throw his if they want the stones to hit the water simultanously, we need to calculate the time needed to hit the water to both rocks independent each other, and just take the difference.

For the rock dropped by Bill, as the only influence on it is gravity (accelerating it downwards with an acceleration equal to g), and v₀ =0, we can use the following kinematic equation:

$y = \frac{1}{2} * g * t^{2}$

where y = height = 40 ft.

As all the parameters are given in SI units, it is  advisable to convert this value to m, as follows:

$y = 40 ft*\frac{0.3048m}{1 ft} = 12.2 m$

Now, we can solve for t, as follows:

$t = \sqrt{\frac{2*y}{g}} = \sqrt{\frac{2*12.2m}{9.8m/s2}} = 1.58 s$

For the rock thrown down at 10 m/s, the kinematic equation we just have used becomes:

$y = v0*t +\frac{1}{2} * g * t^{2}$

This leaves us a quadratic equation on t, as follows:

$t = \frac{-10m/s}{9.8m/s2} +/- \sqrt{10m/s^{2} -4*4.9m/s2*(-12.2m)} = -1.02 s +/- 1.88s$

Taking the positive root, we have:

t = -1.02 s + 1.88 s = 0.86 s

So, in order to get that both rocks hit the water at the same time, Ted will need to wait the difference between both times:

Δt = 0.86 s - 1.58s = -0.72 s