Question

a body is thrown vertically upward with initial velocity of 9.8m/s .it will reach what height?

Answer 1

v^2=u^2+2as
0=9.8^2-2×9.8×s
s=9.8/2=4.8m

Answer 2

Answer:

Explanation:

Solution,

Here, we have

Initial velocity, u = 9.8 m/s

Final velocity, v = 0

Acceleration, a = g = - 9.8 m/s²

To Find,

Height reached , s = ?

According to the 3rd equation of motion,

We know that,

v² - u² = 2gs

So, putting all the values, we get

⇒ v² - u² = 2gs

⇒ (0)² - (9.8)² = 2 × (- 9.8) × s

⇒ 96.04 = - 19.6 s

⇒ 96.04/19.6 = s

⇒ s = 4.9 m.

Hence, the height reached by ball is 4.9 m.