Get Started
1066

Q: A ball is thrown vertically upward from the ground. It crosses a point at the height of 25 m twice at an interval of 4 seconds. The ball was thrown with the velocity of

  • 1
    18 m/s
  • 2
    25 m/s
  • 3
    30 m/s
  • 4
    36 m/s
  • Show Answer
  • Workspace

Answer : 3. "30 m/s"
Explanation :

Answer: C) 30 m/s Explanation: The interval between object pass the same point is 4 sec. That means in 2 sec, object reaches the top and in next 2 sec, it again reaches the same point. By the info given, we can find the velocity of that point by using : v=u+(-gt) 0= u - 10×2 u = 20 Then the velocity at 25m is 20m/sec. so the initial velocity is -v2 = u2 + (-2gh)400 = u2 - 2x 10 x 25u2 = 900 => u = 30 m/s

Previous Next
IIT JEE CAT Bank Exams GATE

The Most Comprehensive Exam Preparation Platform

Get the Examsbook Prep App Today