question_answer

If a ball is thrown vertically upwards with speed\[u,\]the distance covered during the last t second of its ascent is

A)  \[ut-\frac{1}{2}g{{t}^{2}}\]

B)  \[\frac{1}{2}g{{t}^{2}}\]

C)  \[(u+gt)t\]

D)  \[ut\]

Correct Answer: B

Solution :

  When falling from topmost point distance travelled in t sec\[=\frac{1}{2}g{{t}^{2}}\] \[\therefore \]Distance travelled in last t sec of ascent \[=\frac{1}{2}g{{t}^{2}}\]