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/s²?
Let the origin be the ground level, with the upward direction as positive.
For ball A (thrown upward from ground):
Initial position, y₀_A = 0
Initial velocity, u_A = +20 m/s
Equation of motion: y_A(t) = y₀_A + u_A*t – (1/2)gt² = 0 + 20t – (1/2)(9.8)t² = 20t – 4.9t²
For ball B (thrown downward from 40 m):
Initial position, y₀_B = 40 m
Initial velocity, u_B = -20 m/s
Equation of motion: y_B(t) = y₀_B + u_B*t – (1/2)gt² = 40 – 20t – (1/2)(9.8)t² = 40 – 20t – 4.9t²
Collision occurs when y_A(t) = y_B(t):
20t – 4.9t² = 40 – 20t – 4.9t²
Add 20t and 4.9t² to both sides:
40t = 40
t = 1 second
Substitute t = 1s into either equation to find the height:
y_A(1) = 20(1) – 4.9(1)² = 20 – 4.9 = 15.1 m
y_B(1) = 40 – 20(1) – 4.9(1)² = 40 – 20 – 4.9 = 15.1 m
The balls collide after 1 second at a height of 15.1 m from the ground.
– Define a consistent coordinate system (origin and positive direction).
– Set the positions of the two objects equal to find the time of collision.
– Use the time of collision to find the position (height) of collision.
– The relative velocity approach could also be used for the time of collision: v_rel = v_A – v_B. Initial relative velocity = 20 – (-20) = 40 m/s. The relative acceleration is g – g = 0. The initial separation is 40m. Time to collide = separation / relative velocity = 40m / 40m/s = 1s. This confirms the time calculation.