Question #262259

A stone is thrown straight downward with initial speed 8.0 m/s from a height of 25 m. Find (a) the time it takes to reach the ground and (b) the speed with which it strikes.

1
Expert's answer
2021-11-07T19:25:44-0500

(a) Let's take the downwards as the positive direction. We can find the time that the stone takes to reach the ground from the kinematic equation:


d=v0t+12gt2,d=v_0t+\dfrac{1}{2}gt^2,25=8t+4.9t2,25=8t+4.9t^2,4.9t2+8t25=0.4.9t^2+8t-25=0.

This quadratic equation has two roots: t1=1.59 st_1=1.59\ s and t2=3.22 s.t_2=-3.22\ s. Since time can't be negative the correct answer is t=1.59 st=1.59\ s.

(b) We can find the speed with which the stone strikes the ground from the kinematic equation:


v=v0+gt=8 ms+9.8 ms2×1.59 s=23.6 ms.v=v_0+gt=8\ \dfrac{m}{s}+9.8\ \dfrac{m}{s^2}\times1.59\ s=23.6\ \dfrac{m}{s}.

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