A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 55.0 m high (Fig. 2-34).

A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 55.0 m high (Fig. 2-34).



Figure 2-34



(a) How much later does it reach the bottom of the cliff?

s

(b) What is its speed just before hitting?

m/s

(c) What total distance did it travel?

m

A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 55.0 m high (Fig. 2-34).
hey there.

to solve the problem we'll use the classic formula:



y = x0 + v0 * t + 1/2 * a * (t^2)



when we put in

y0 = 55 m

v0 = 12 m/s

a = -9.8 m/sec^2 (as the stone is free falling)

demanding that y=0 we get the following quadratic equation:



4.9* (t^2) - 12*t -55 = 0



the positive solution for that equation is t = 4.79 sec, and that is the answer to part (a)



to find the speed it hits the groun we'll use



v = v0 + at



putting in v0 = 12 m/s , a = -9.8 m/(sec^2) and t = 4.79 sec (we just found that) we get that



v = 12 - 9.8 * 4.79 = -34.942 m/sec (the minus sign means that the direction is opposite to the direction of the axis we chose. I chose the axis to be pointing up, so the final velocity is pointing down. makes sense, doesn't it?)





(c) hmmmmm interesting part. let's calculate how much time did it take from the throw till the stone stopped for a brief moment in mid air:



0 = v0 - g*t --%26gt; t = v0 / g = 12 m/s / 9.8 m/(sec^2) = 1.2244 sec.

lets see how high the stone got (can stones get high????) when it stopped (the highest point in it's journey):



y = v0 * t -1/2 * g * (t^2) = 12 * 1.2244 - 1/2 * 9.8 * (1.2244^2) = 7.347 m



the total distance the stone traveled is twice that distance (one time upwards and another downwards) plus the height of the cliff



total distance = 2*7.347 + 55 = 69.69 m

games hardware