A ball is thrown vertically upward with a speed of 20 m per second from the top of a tower 60 m height the time taken to reach the ground days
Solution
Given:
- Initial velocity: u = +20 m/s (upward)
- Height of tower: h = 60 m
- Acceleration: g = −10 m/s² (downward)
- Taking upward as positive, displacement when ball hits ground = −60 m
Using the equation of motion:
$$s = ut + \frac{1}{2}at^2$$
$$-60 = 20t + \frac{1}{2}(-10)t^2$$
$$-60 = 20t - 5t^2$$
$$5t^2 - 20t - 60 = 0$$
$$t^2 - 4t - 12 = 0$$
(t − 6)(t + 2) = 0
$$t = 6 \text{ s} \quad \text{or} \quad t = -2 \text{ s (rejected)}$$
∴ Time taken to reach the ground = 6 seconds
Quick sanity check:
- Ball rises for: t_up = u/g = 20/10 = 2 s
- Max height above ground = 60 + (20²/20) = 60 + 20 = 80 m
- Then falls 80 m from rest: 80 = ½(10)t² → t = 4 s
- Total = 2 + 4 = 6 s ✓
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