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.

Expert's answer

(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}.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

LATEST TUTORIALS
APPROVED BY CLIENTS