Question
Question: An object is thrown upward with a velocity u, then the displacement-time graph is? 

Solution
The question is based on one dimensional motion of a particle. As here an object is thrown vertically upward and we consider the one dimensional motion of the object in vertical direction. As there is gravity and it attracts the object towards the earth surface hence the velocity of the object is affected by gravity.
Complete step by step solution:
Step 1: Let us consider an object is thrown vertically upward with a velocity u. And after time t the object reaches at height s. Then the one dimensional equation of motion is,
s=ut−21gt2---------------------- (1) 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.
