As the upward direction is taken as positive, the velocity at t=0 is u=−100ms−1.
As it is falling under gravity, its acceleration is constant and a=−10ms−2.
Applying equation of motion for constant acceleration during the time of fall i.e., for 0 to 10s,
v=u+at⇒v=−100−10×10⇒v=−200ms−1
After 10s it hits the ground and remains at rest.
Therefore, option A is the correct representation.



