Question 7 of 11advanced🔧 ApplyNumerical5 marks

A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t=0t = 0 to 12 s.

Correct Answer

Given: Ball dropped from height h = 90 m, g = 10 m/s², loses 1/10 speed at each collision.

Step 1: First fall (t = 0 to t₁) Using s = ut + ½gt²: 90 = 0 + 5t₁² t₁ = 32\sqrt{2} s ≈\approx 4.24 s Speed at impact: v₁ = gt₁ = 302\sqrt{2} m/s ≈\approx 42.43 m/s

Step 2: After first collision Speed = (9/10) × 302\sqrt{2} = 272\sqrt{2} m/s ≈\approx 38.18 m/s

Step 3: Upward motion Time to max height = 272\sqrt{2}/10 = 2.72\sqrt{2} s ≈\approx 3.82 s Time at max height = 32\sqrt{2} + 2.72\sqrt{2} = 5.72\sqrt{2} s ≈\approx 8.06 s

Step 4: Downward motion Time to fall back = 2.72\sqrt{2} s Time at second impact = 5.72\sqrt{2} + 2.72\sqrt{2} = 8.42\sqrt{2} s ≈\approx 11.88 s

Step 5: After second collision Speed = (9/10) × 272\sqrt{2} = 24.32\sqrt{2} m/s ≈\approx 34.37 m/s

Step 6: At t = 12 s Speed = 24.32\sqrt{2} - 10×0.1210 \times 0.12 ≈\approx 33.17 m/s

Speed-Time Graph Description: The graph shows linear segments with slope ±g:

  • Line from (0, 0) to (4.24, 42.43) — ball falling
  • Vertical drop at t = 4.24 s from 42.43 to 38.18 m/s — collision
  • Line from (4.24, 38.18) to (8.06, 0) — ball rising
  • Line from (8.06, 0) to (11.88, 38.18) — ball falling
  • Vertical drop at t = 11.88 s from 38.18 to 34.37 m/s — collision
  • Line from (11.88, 34.37) to (12, 33.17) — ball rising

The graph forms triangular peaks with decreasing heights due to energy loss at each collision.

Exercise: EXERCISES | Q: 2.8 | (Page 13)
For More Understanding

Explanation

This problem applies equations of motion for free fall. The ball undergoes multiple cycles of falling and rising, with speed reducing by 1/10 at each floor collision. Using s = ut + ½gt² and v = u + at, we calculate key time points and speeds. The speed-time graph shows characteristic triangular peaks with vertical drops at collision points, representing instantaneous speed loss. The context from Example 2.3 demonstrates similar methodology for calculating time and velocity in vertical motion problems.

Solution Steps

  1. Step 1: Calculate first fall time using s = ut + ½gt² → t₁ = 32\sqrt{2} s ≈\approx 4.24 s

  2. Step 2: Calculate speed at first impact using v = gt → v₁ = 302\sqrt{2} m/s ≈\approx 42.43 m/s

  3. Step 3: Calculate speed after first collision → v₁' = 272\sqrt{2} m/s ≈\approx 38.18 m/s

  4. Step 4: Calculate time to reach max height → t_up = 2.72\sqrt{2} s ≈\approx 3.82 s

  5. Step 5: Calculate time at max height → t_max = 5.72\sqrt{2} s ≈\approx 8.06 s

  6. Step 6: Calculate time at second impact → t₂ = 8.42\sqrt{2} s ≈\approx 11.88 s

  7. Step 7: Calculate speed after second collision → v₂' = 24.32\sqrt{2} m/s ≈\approx 34.37 m/s

  8. Step 8: Calculate speed at t = 12 s → v ≈\approx 33.17 m/s

  9. Step 9: Plot speed-time graph with calculated points

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