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Question: An object is thrown upward with a velocity u, then the displacement-time graph is? ![](https://www...

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

Explanation

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 uu. And after time tt the object reaches at height ss. Then the one dimensional equation of motion is,
s=ut12gt2s = ut - \dfrac{1}{2}g{t^2}---------------------- (1) where, g=g = gravitational constant.
Step 2: From equation (1) we can see that vertical distance hhis a quadratic function of t.
Now, the maximum height of the object is obtained from the relation v2=u22gs{v^2} = {u^2} - 2gs---- (2)
Where, vv= the velocity of the object at heightsmax{s_{\max }}. smax{s_{\max }} is the maximum height of the object.
At maximum height v=0v = 0that means the object stops atsmax{s_{\max }}.
Then from equation (2) we get 0=u22gsmax0 = {u^2} - 2g{s_{\max }}
Or, u2=2gsmax{u^2} = 2g{s_{\max }} Or, smax=u22g{s_{\max }} = \dfrac{{{u^2}}}{{2g}}
Step 3: Now if the speed of the object is vv at timett. Then we get, v=ugtv = u - gt----- (3)
At height smax{s_{\max }} , v=0v = 0and t=tmaxt = {t_{\max }}; then from equation (3) we get 0=ugtmax0 = u - g{t_{\max }}
Or, tmax=ug{t_{\max }} = \dfrac{u}{g}
Therefore, time taken tg{t_g}to reach ground is twice of tmax{t_{\max }}. That is tg=2ug{t_g} = \dfrac{{2u}}{g}.

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 accelerationgg. Let an object be thrown vertically upward with an initial speeduu. And at time tt the velocity of the object will be vv and height will be ssthen three equations of motion are:
v=ugtv = u - gt
s=ut12gt2s = ut - \dfrac{1}{2}g{t^2}
v2=u22gs{v^2} = {u^2} - 2gs.