A ball is thrown vertically upward from the ground and passes a point 25 m above the ground twice, the two crossings being 4 s apart. What was the initial speed of projection? (Take g ≈ 10 m/s^2.)

Introduction / Context:
Vertical motion under gravity is a standard topic in kinematics. When a ball is thrown straight up, it passes the same height twice, once while going up and once while coming down. The time gap between these two events can be related to the initial speed. This question uses that idea to find the initial speed of projection from the given height and time interval.

Given Data / Assumptions:

• Initial velocity of the ball is u (upwards).
• Acceleration due to gravity g is taken as 10 m/s^2 downward.
• The ball passes height h = 25 m above the ground twice.
• The time between the two crossings of height 25 m is 4 s.
• Motion is vertical with no air resistance.

Concept / Approach:
For vertical motion with constant acceleration, the displacement y at time t is given by y = u t - (1/2) g t^2 when upward is taken as positive. When the ball passes a given height h twice, the corresponding times t1 and t2 are roots of a quadratic equation. For such a quadratic, the sum and difference of roots have simple relations. The time gap t2 - t1 depends only on u, h, and g. By using this relation along with the known interval, we can solve for u without first finding t1 or t2 individually.

Step-by-Step Solution:
Step 1: Write the equation for height h = 25 m: 25 = u t - (1/2) g t^2. Step 2: Rearrange to standard form: (1/2) g t^2 - u t + 25 = 0. With g = 10 m/s^2, this becomes 5 t^2 - u t + 25 = 0. Step 3: The two roots of this quadratic are t1 and t2, the times when the ball is at 25 m. For a quadratic a t^2 + b t + c = 0, the difference between roots is t2 - t1 = sqrt(b^2 - 4 a c) / a. Step 4: Here, a = 5, b = -u, and c = 25. The time interval between the two crossings is given as 4 s, so 4 = sqrt(b^2 - 4 a c) / a. Step 5: Substitute a and c: 4 = sqrt(u^2 - 4 * 5 * 25) / 5 = sqrt(u^2 - 500) / 5. Step 6: Multiply both sides by 5: 20 = sqrt(u^2 - 500). Square both sides: 400 = u^2 - 500, so u^2 = 900. Step 7: Therefore u = 30 m/s, taking the positive root because the ball is thrown upward.

Verification / Alternative check:
We can verify quickly by computing the times with u = 30 m/s. Use 25 = 30 t - 5 t^2. Rewrite as 5 t^2 - 30 t + 25 = 0. Solving gives t = [30 ± sqrt(900 - 500)] / 10 = [30 ± sqrt(400)] / 10 = (30 ± 20) / 10, so t1 = 1 s and t2 = 5 s. The time gap is 5 - 1 = 4 s, which matches the given condition. This confirms that u = 30 m/s is correct.

Why Other Options Are Wrong:
18 m/s: For this speed, the maximum height itself would be too small to pass 25 m twice, so no real solution exists. 25 m/s: Solving the quadratic with this value leads to a different time gap than 4 s. 36 m/s: This would produce a time interval greater than 4 s between the two crossings of 25 m.

Common Pitfalls:
One common mistake is to treat 4 s as the total time of flight rather than the interval between two specific heights. Another is to try to find t1 and t2 separately without using the quadratic root relations, which can lead to algebra errors. Remember that when a projectile passes the same height twice, using the sum and difference of roots is an efficient and elegant method.

Final Answer:
The initial speed of projection was 30 m/s.