Question

Question #b1245

Answer 1

`y(t)= 1/2 g t - 1/2 g t^2`

Explanation:

Using Equation of motion:
If g is taken to be the acceleration due to gravity, its direction being positive when an object is traveling downwards, the equation of motion, giving the height of the object, in the given case would be `y(t)=ut-1/2 g t^2`, where u is the initial velocity.

Initial velocity is given to be `g/2`. Thus the required equation would be `y(t)= 1/2 g t - 1/2 g t^2`

Using calculus
If y represents the distance traveled in time t , then velocity of the object would be `dy/dt` and the acceleration would be given by `(d^2y)/dt^2`

Since the object is moving upwards under force of gravity, the acceleration due to gravity would be -g.

Thus `(d^2y)/dt^2 =-g`. Integrating it would give dy/dt= -gt +C1. Since at t=0, velocity is `g/2= dy/dt`. This would give C1=`g/2` and hence `dy/dt= -g t +g/2` .

Integrating once again would give `y(t)= 1/2 g t- 1/2 g t^2 +C2`. Applying the initial condition that at t=0, distance traveled y(t)=0, gives C2 =0.

The required equation is thus `y(t)= 1/2 g t - 1/2 g t^2`