Question

A ball is thrown up vertically with an initial speed of 20ms.when it has reached 3÷4 of the maximum height,speed of that ball is (take,g=10 ms)

Answer 1

Answer:

10 m/s

Explanation:

Maximum height = u²/(2g)

= (20 m/s)²/(2 × 10 m/s²)

= (400 / 20) m

= 20 m

3/4 of 20 m = 15 m

Velocity after reaching 15 m height

v = sqrt(u² + 2aS)

= sqrt((20 m/s)² + (2 × -10 m/s² × 15 m))

= sqrt(400 - 300) m/s

= sqrt(100) m/s

= 10 m/s