Q25 / 72 · NTA Abhyas (JEE Main) 2020 · Physics · Motion In Two Dimensions

A ball is projected from the ground at an angle θ with the horizontal. After 1 s it is moving at an angle 45° with the horizontal and after 2 s, it is moving horizontally. What is the velocity of projection of the ball?

  • A
    103 m s-1
  • B
    203 m s-1
  • Correct answer
    C
    105 m s-1
  • D
    202 m s-1

Solution

The ball will be vertical at the highest point so

tup= usinθg=2 ⇒usinθ=20   .. …….(1)

∵ tanα=usinθ-gtu cosθ

⇒1=20-g(1)ucosθ⇒ucosθ=10… . .(2)

Squaring and adding equation (1) & equation (2), we get

usinθ2+ucosθ2=202+102

⇒   u2=500

⇒ u=105 m s-1

Hence, the correct answer is option: 'C'.

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