Problem: A ball is projected vertically upward it's speed at half of maximum height is 20m/sec. The maximum height attained by it is (take g=10ms^2)
a) 35m b) 15m C) 25m d) 40m?

Solution:
Given,
Initial velocity (u) = ?
Final velocity (v) = 0 (As the ball reaches maximum height, the velocity becomes zero)
Acceleration (a) = g = 10m/s²
Using the formula, v² - u² = 2as
0² - u² = 2 × 10 × s
u = √(20s) ---(1)

Let the maximum height attained by the ball be H.
At half of maximum height, the velocity of the ball is 20m/s.
Using the formula, v² - u² = 2as
(20)² - (u)² = 2 × 10 × (H/2)
400 - u² = 10H ---(2)

Substituting the value of u from equation (1) in equation (2), we get
400 - 20s = 10H
H = 40 - 2s

Therefore, the maximum height attained by the ball is 40 - 2s meters. To find the value of s, we can use the fact that at the maximum height, the velocity of the ball is zero.
Using the formula, v = u + at
0 = √(20s) - 10t
t = √(2s)/2

Substituting the value of t in the equation for maximum height, we get
H = 40 - 2s = 40 - s√2

Therefore, the maximum height attained by the ball is 40 - s√2 meters.
Now, to find the value of s, we need to use the fact that at half of the maximum height, the velocity of the ball is 20m/s.
Using the formula, v = u + at
20 = √(20s) - 10t
t = (2 + √2)/2

Substituting the value of t in the equation for velocity at half of maximum height, we get
20 = √(20s) - 10(2 + √2)/2
s = 25

Therefore, the maximum height attained by the ball is 40 - s√2 = 40 - 5√2 = 25 meters.

Answer: The maximum height attained by the ball is 25m (Option C).