Question

Why is the value of 'g' taken as negative when a body is thrown vertically upwards?​

Answer 1

See below

Step-by-step explanation:

`\displaystyle {\textsf {Acceleration is the rate of change of velocity with time:}}`

`\boxed {g = (dv)/(dt)}`

`\sf {where \;dv = \;change \;in \;velocity\; and \;dt \;the \;change \;in \;time}`

here g is the gravitational force

When a body is thrown vertically upward, it has an initial speed. As it keeps going up, it speed keeps decreasing until it becomes zero, then the ball starts dropping down

On the upward movement, the change in velocity is negative while the change in time is always positive

So dv/dt = a negative number divided by a positive number which is a negative value for g in the equation for g