Question

Two students are on a balcony 18.7 m above the street. One student throws a ball (ball 1) vertically downward at 14.6 m/s; at the same instant, the other student 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 magnitude m/s direction ball 2 magnitude m/s direction (c) How far apart are the balls 0.500 s after they are thrown? m

Answer #1

here,

height , h0 = 18.7 m

intial speed of 1, u1 = 14.6 m/s

initial speed of 2 , u2 = - 14.6 m/s

a)

let the time taken to rach ground be t

h0 = u2 * t + 0.5 * g * t^2

18.7 = - 14.6 * t + 0.5 * 9.81 * t^2

solving for t

t = -0.97 s and 3.94 s

the time taken for ball 1 is 0.97 s

the time taken for ball 2 is 3.94 s

b)

the final speeds of ball is same at the ground and points downwards

the magnitude of final velocity , v = u1 + g * t1

v = 14.6 + 9.81 * 0.97 m/s

v = 24.1 m/s

c)

at t1 = 0.5 s

h1 = u1 * t + 0.5 * g * t^2

h1 = 14.6 * 0.5 + 0.5 * 9.81 * 0.5^2

h1 = 8.52 m

for seccond ball

h2 = u2 * t + 0.5 * g * t^2

h2 = - 14.6 * 0.5 + 0.5 * 9.81 * 0.5^2

h2 = - 6.08 m

the height difference , h = h1 - h2

h = 14.6 m

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