# A = body Thrown up with velocity of 20 metre per second find the time it takes to reach the maximum height given g= 10 metre per second square

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For the projectile motion, the maximum height attained by an object is given as:

`h_max = (v^2 sin^2theta)/(2g)`

where, hmax is the maximum height attained during the projectile motion

v: initial velocity = 20 m/sec

g: acceleration due to gravity (= 10 m/sec^2), and

`theta` : angle that the projectile motion makes with the horizontal: 90 degrees in this case (as the body is 'thrown up')

and the time taken to reach maximum height is given as :

`t_h_max = (v sin theta)/g`

thus, `h_max = [(20)^2 xx sin^2 (pi/2)]/(2xx10) = 20 m`

and, `t_h_max = (20 xx sin (pi/2))/10 = 2 sec`

Thus, it will take **2 sec** to reach the maximum height of 20 m.

Hope this helps

The speed of the vertical motion is calculated by the following expression:

v = vi ± gt

v → is the velocity at any instant of movement.

vi → is the initial velocity.

g → is the acceleration of gravity.

t → is the time.

We use the positive sign for downward movement and the negative sign for upward movement. Since the body stops at the top(Hmax), we have that in our equation v = 0, so that:

0 = vi – gt

t = vi/g = 20(m/s)/10(m/s^2) = 2 s

**Then it takes 2 seconds to reach the maximum height**.