Q45 / 72 · NTA Abhyas (JEE Main) 2020 · Physics · Motion In Two Dimensions

A ball is projected from point A with a velocity, 10 m s-1 perpendicular to the inclined plane as shown in the figure. Range of the ball on the inclined plane is :

Figure

  • Correct answer
    A
    4 0 3 m
  • B
    2 0 13 m
  • C
    13 20 m
  • D
    13 40 m

Solution

T = 2u sin 90 o g cos 30 o = 2 u g cos 3 0 o

R = 1 2 × g sin 30 o ×  4 u 2 g 2 cos 2 30 o

R = 2 u 2 × 4 2 × 1 0 × 3 = 2 × 1 0 × 1 0 × 4 2 × 1 0 × 3 = 4 0 3 m

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

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