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
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.