11. একটি বস্তুকে অনুভূমিকের 60° কোণে K গতিশক্তি দিয়ে ছোঁড়া হল। গতিপথের সর্বোচ্চ বিন্দুতে এর গতিশক্তি – (An object is thrown at a 60 ° angle of horizontal with K kinetic force. Its kinetic energy at the highest point of the track-)
| (a) K, (b) শূন্য (০), (c) K/4 (d) K/2
which one is the answer and how ?
Answers
Answer:
At the highest point of motion ,the vertical component of velocity becomes zero. Only horizontal component of velocity exists and it remains constant through out the motion(if air resistance is neglected).
As kinetic energy is given, it's velocity can be easily derived.
E = (1/2)×M×v^2.
=> V= √(2E/M).
The horizontal component of velocity will be(given angle of projection 60°) vcos60°.
=> V (horizontal) = √(2E/M)×(1/2) = √(E/M×1/2).
=> K.E at heighest point = 1/2×M× v ^2 .( Here v is horizontal velocity).
=> K.E = E/4.
FOLLOW ME
Explanation:
Answer:
At the highest point of motion ,the vertical component of velocity becomes zero. Only horizontal component of velocity exists and it remains constant through out the motion(if air resistance is neglected).
As kinetic energy is given, it's velocity can be easily derived.
E = (1/2)×M×v^2.
=> V= √(2E/M).
The horizontal component of velocity will be(given angle of projection 60°) vcos60°.
=> V (horizontal) = √(2E/M)×(1/2) = √(E/M×1/2).
=> K.E at heighest point = 1/2×M× v ^2 .( Here v is horizontal velocity).
=> K.E = E/4.
Explanation: