Question

In the problem "up" is taken positive
A ball is thrown vertically upward with an initial velocity 19.6m/s after 5s has a velocity of

Answer 1

Explanation:

Max height attained by it is

Hmax = u^2 = 19.6 × 19.6 ÷ 2 × 9.8

2g

Hmax = 19.6m

putting the value in 3rd Eq. of motion

v^2 - u^2 = 2as

v^2 = 19.6 × 19.6 + 2 × 9.8 × 19.6

v^2 = 384.16 + 384.16

v^2 = 768.32

v = 27.71 m/s or 28 m/s

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