Question

A body is projected vertically upward from 19.6 metre per second with its initial velocity what is its height from the ground

Answer 1

Answer:

19.6 m (approx)

Explanation:

initial velocity (u) = 19.6 m/s

final velocity (v) = 0 m/s [As the body stops at a certain height]

acceleration(a) = -g [Acceleration due to gravity]

[Acceleration is negative as the object slows down as it is going against the gravity]

Let the height be s

Therefore

Using 3rd equation of motion

$v^{2} = u^{2} + 2as\\0^{2} = 19.6^{2} + 2*(-g)*s\\0 = 384.16 + 2*(-9.8)*s \ ......... [g = 9.8 m/s^{2}]\\-384.16 = -19.6*s\\s = \frac{-384.16}{-19.6} = 19.6 m$

Therefore

Height = 19.6 m(approx)

Answer 2

Given that, a body is projected vertically upward from 19.6 metre per second with its initial velocity.

So, Initial velocity (u) of the body is 19.6 m/s

Final velocity (v) of the body is 0 m/s

Acceleration due to gravity (a) is -9.8 m/s²

We have to find the height (s) of the body from the ground.

Using the Third Equation of Motion

v² - u² = 2as

Substitute the known values

(0)² - (19.6)² = 2 × (-9.8) × s

0 - 384.16 = -19.6s

s= 384.16/19.6

s= 19.6

Therefore, the height of the body from the ground is 19.6 m.