Question

A man stands at the edge of a cliff and throws a rock downward with a speed of 12.0 m/* . Sometime later it strikes the ground 110 m below the place where it was thrown. (a) How long does it take to reach the ground? (b) What is the speed of the rock at impact?

Answer 1

(a)The problem is illustrated in Fig. Since the rock is thrown downward, the initial velocity of the rock is v0 = −12.0 m/*, and of course a = −9.80 m/s2 . When the rock hits the ground its y coordinate is y = −110 m, so in this part we are asking “At what time does y = −110 m?” y is given by

which is a quadratic equation.Using the quadratic formula, there are two possible answers, given by

A little calculator work gives the two (?)

So which is the answer? (There can only be one time of impact!) The answer must be the second one because a negative time t is meaningless; the rock was thrown at t = 0. Therefore the rock takes 3.67 * to reach the ground.

(b) We need to find the velocity of the rock at the the time found in part (a). The velocity of the rock is given by

and so the speed of the rock at impact is 18.1 m/* .