Two balls, A and B, are thrown simultaneously. A vertically upward with a speed of 20 m\/s from the ground and B vertically downward from a height of 40 m with the same speed and along the same line of motion. At what points do the two balls collide by taking acceleration due to gravity as 9.8 m\/s\u00b2?<\/p>\n
[amp_mcq option1=”The balls will collide after 3s at a height of 30\u00b72 m from the ground” option2=”The balls will collide after 2s at a height of 20\u00b71 m from the ground” option3=”The balls will collide after 1s at a height of 15\u00b71 m from the ground” option4=”The balls will collide after 5s at a height of 20 m from the ground” correct=”option3″]<\/p>\n
For ball B (thrown downward from 40 m):
\nInitial position, y\u2080_B = 40 m
\nInitial velocity, u_B = -20 m\/s
\nEquation of motion: y_B(t) = y\u2080_B + u_B*t – (1\/2)gt\u00b2 = 40 – 20t – (1\/2)(9.8)t\u00b2 = 40 – 20t – 4.9t\u00b2<\/p>\n
Collision occurs when y_A(t) = y_B(t):
\n20t – 4.9t\u00b2 = 40 – 20t – 4.9t\u00b2
\nAdd 20t and 4.9t\u00b2 to both sides:
\n40t = 40
\nt = 1 second<\/p>\n
Substitute t = 1s into either equation to find the height:
\ny_A(1) = 20(1) – 4.9(1)\u00b2 = 20 – 4.9 = 15.1 m
\ny_B(1) = 40 – 20(1) – 4.9(1)\u00b2 = 40 – 20 – 4.9 = 15.1 m<\/p>\n