A ball is thrown vertically upwards with a velocity of 19.6 ms⁻¹ from the top of a tower. The ball strikes the ground after 6 s. The height from the ground up to which the ball can rise will be (k/5) m. The value of k is __________. (use g=9.8 m / s²)

Solution

Physics sir · 3 steps

Use the 6 s flight to find the tower's height, then add the extra rise above the top.

1. Height of tower. Take upward as positive, with the origin at the top of the tower. The ball ends below the start, so its displacement in 6 s is -h, where h is the tower's height.

   -h = ut - 1/2gt² = 19.6 × 6 - 1/2 × 9.8 × 6² = 117.6 - 176.4 = -58.8 m

2. Rise above the top. The ball keeps rising until its velocity is zero. Using v² = u² - 2gH gives H = u²/2g = 19.6²/(2 × 9.8) = 19.6 m above the tower.
3. So. The greatest height above the ground is 58.8 + 19.6 = 78.4 m. Setting k/5 = 78.4 gives k = 392.

The common slip: 98. Taking only the rise above the tower, 19.6 m, as the height gives k = 98. You must add the 58.8 m of the tower itself.

The idea to keep. Measure every position from the launch point with one sign convention. Then a single displacement equation covers the whole flight, the upward part and the fall together.

Answer from the NTA final answer key. Solution by PrepWiser.