Question
Question: An object is thrown upward with a velocity u, then the displacement-time graph is?  where, g= gravitational constant.
Step 2: From equation (1) we can see that vertical distance his a quadratic function of t.
Now, the maximum height of the object is obtained from the relation v2=u2−2gs---- (2)
Where, v= the velocity of the object at heightsmax. smax is the maximum height of the object.
At maximum height v=0that means the object stops atsmax.
Then from equation (2) we get 0=u2−2gsmax
Or, u2=2gsmax Or, smax=2gu2
Step 3: Now if the speed of the object is v at timet. Then we get, v=u−gt----- (3)
At height smax , v=0and t=tmax; then from equation (3) we get 0=u−gtmax
Or, tmax=gu
Therefore, time taken tgto reach ground is twice of tmax. That is tg=g2u.
Step 4: As per explanation given above the displacement – time graph is represented as
Therefore, option (A) is correct.
Note: Student has to remember the equation of one dimensional motion under the gravitational accelerationg. Let an object be thrown vertically upward with an initial speedu. And at time t the velocity of the object will be v and height will be sthen three equations of motion are:
v=u−gt
s=ut−21gt2
v2=u2−2gs.