Two students are on a balcony 20.0 m above the street. One student throws aball

Question

Two students are on a balcony 20.0 m above the street. One student throws aball (ball 1) vertically downward at 11.1 m/s; at the same instant, the otherstudent throws a ball (ball 2) vertically upward at the same speed.The second ball just misses the balcony on the way down. (a) What is the difference in the two ball's time in the air?
s

(b) What is the velocity of each ball as it strikes the ground? ball 1 m/s ball 2 m/s
(c) How far apart are the balls 0.800 s after they are thrown?
m (a) What is the difference in the two ball's time in the air?
s

(b) What is the velocity of each ball as it strikes the ground? ball 1 m/s ball 2 m/s
(c) How far apart are the balls 0.800 s after they are thrown?
m ball 1 m/s ball 2 m/s

Explanation / Answer

The height of the balcony above the ground is S = 20.0 m The first student throws the ball 1 vertically downward atu1= 11.1 m/s The other student throws a ball 2 vertically upward atu2= 11.1 m/s The second ball just misses the balcony on the way down. (a)The time taken by the first ball to strike the groundis v1= u1 + gt Here,v1= 0 m/s or t = -(u1/-g) = (11.1/9.8) = 1.13 s The height reached by the ball 2 above the ground is v22 - u22 =2gH Here,v2= 0 m/s or (0)2 - u22 = 2gH or H = -(u22/2g) or H = -((11.1)2/2 * 9.8) = -6.3 m The negative value of height indicates that the ball 2 movesagainst the force of gravity. The time spend by the ball 2 in air is S = u2t+ (1/2)gt2 Here,S = (20.0 + 6.3) m = 26.3 m and g = 9.8m/s2 or 11.1t + (1/2) * 9.8 * t2 = 26.3 or 4.9t2 + 11.1t - 26.3 = 0 or t1= 1.44 s and t2= -3.71 s Therefore,the difference in the two ball's time in the airis t = (t1 - t) = (1.44 - 1.13) s = 0.31 s (b)The velocity of the ball 1 as it strikes the groundis v12 - u12 =2gS or v12 = u12 +2gS or v12 = (11.1)2 + 2 * (-9.8)* 20.0 = 268.79 or v1 = 16.4 m/s Similarly,the velocity of the ball 2 as it strikes theground is v22 - u22 =2gS or v22 = u22 +2gS Here,S = 26.3 m or v22 = (11.1)2 + 2 * (-9.8)* 26.3 = 392.27 or v2 = 19.8 m/s (c)The height reached by ball 1 when t = 0.800 s is S = u1t + (1/2)gt2 or S = 11.1 * 0.800 + (1/2) * (-9.8) *(0.800)2 or S = -5.74 m The height reached by ball 2 when t = 0.800 s is S = u2t + (1/2)gt2 or S = 11.1 * 0.800 + (1/2) * 9.8 * (0.800)2 or S = 12.01 m Therefore,the balls are (12.01 - 5.74) m = 6.27 m apart after0.800 s when they are thrown. or S = 12.01 m Therefore,the balls are (12.01 - 5.74) m = 6.27 m apart after0.800 s when they are thrown.

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