Question

A ball is thrown up at 4 metre per second find maximum height reached by the ball​

Answer 1

we have,

H=u^2/2g=4×4/2×10=0.8m

Answer 2

Explanation:

Let the maximum height be s.

Therefore by 3rd equation of motion

${v}^{2} = {u}^{2} + 2as \\ = {0}^{2} = {4}^{2} + 2 \times ( - 9.8) \times s \\ = 0 = 16 - 19.6s \\ = - 16 = - 19.6s \\ = s = - 16 \div - 19.6 \\ = s = 0.816metres$

Hope this helps you understand my friend ❤️ ❤️ ❤️