A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. (a) At what time after being ejected is the boulder moving at 20.0 m/s upward? (b) At what time is it moving at 20.0 m/s downward? (c) When is the displacement of the boulder from its initial position zero? (d) When is the velocity of the boulder zero? (e) What are the magnitude and direction of the acceleration while the boulder is (i) moving upward? (ii) Moving downward? (iii) At the highest point? (f) Sketch ay-t, vy-t, and y-t graphs for the motion.
Solution 48 E Step 1 : Considering the data given Initial velocity of the boulder v = 40i0 m/s We need to find the time by the boulder to move upward Final velocity v = f0.0 m/s The acceleration acting upon the boulder is given by gravitational force Hence we have a = g = 9.8 m/s 2 Since the boulder is moving against gravitational force Thus we get g = 9.8 m/s 2 Step 2 : We shall use condition v f v + it Rearranging the equation to find t We get vfvi t = g Substituting the values t = 20 m/s40 m/s 9.8 m/s 20 m/s t = 9.8 m/s2 t = 2.04 s Hence it takes 2.04s for a boulder to travel velocity 20m/s upwards Step 3 : We need to find the time by the boulder to move downward With velocity v = 20.0 m/s Since the boulder is moving down its final velocity will be v = 0 m/s f It was ejected with initial velocity v = 40 mis Hence we have time taken as t = vfvi g Substituting the values t = 040 m/2 9.8 m/s 40 m/s t = 9.8 m/s t = 4.08 s Hence it takes 4.08 s for a boulder to travel velocity 20m/s downward Step 4: We need to find the displacement of the boulder from its initial position zero Let us consider the distance travelled by the boulder upwards The average distance is given by v +v (v ) = f i avg up 2 Substituting the values we get (v avg up = 0+40= 20 m/s 2 Distance travelled is given by d up= (v avg)up time taken d up= 20m/s × 2.04 s = 40.8 m Hence the boulder travels 40.8 mupwards in time 2.04 s Similarly Let us consider the distance travelled by the boulder downward The average distance is given by vf+vi (vavg down = 2 Substituting the values we get (v ) = 0+20 = 10 m/s avg down 2 Distance travelled is given by d down = (vavg)down × time taken d down = 10 m/s × 4.08 s = 40.8 m Hence the boulder travels 40.8 mdownward in time 4.08 s The displacement of the boulder from its initial position zero is obtained as d upd down Substituting the values 40.8 m + 40.8 m = 81.6 m Hence the displacement of the boulder is 81.6 m Step 5 : We need to find when will be the velocity of the boulder be zero The velocity of the boulder will zero when it time t = 4.08 s when the boulder is at maximum height