. A particle is projected with a velocity of 20 ms - at an angle of

1.	. A particle is projected with a velocity of 20 ms - at an angle of 60° to the horizontal. The particle hits the horizontal planeagain during its journey. What will be the time of impact?(a) 3.53 s.(b) 2.4 s(C) 1.7 s(d) 1s
Answer» ong>ANSWER: VELOCITY at max height =20cos60 =10 m/s. i.e., using MOMENTUM CONSERVATION 10×2m=m×v ⇒v=20 m/s. Explanation:

. A particle is projected with a velocity of 20 ms - at an angle of 60° to the horizontal. The particle hits the horizontal planeagain during its journey. What will be the time of impact?(a) 3.53 s.(b) 2.4 s(C) 1.7 s(d) 1s

Answer»

ong>ANSWER:

VELOCITY at max height =20cos60

=10 m/s.

i.e.,

using MOMENTUM CONSERVATION

10×2m=m×v

⇒v=20 m/s.

Explanation:

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. A particle is projected with a velocity of 20 ms - at an angle of 60° to the horizontal. The particle hits the horizontal planeagain during its journey. What will be the time of impact?(a) 3.53 s.(b) 2.4 s(C) 1.7 s(d) 1s​

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. A particle is projected with a velocity of 20 ms - at an angle of 60° to the horizontal. The particle hits the horizontal planeagain during its journey. What will be the time of impact?(a) 3.53 s.(b) 2.4 s(C) 1.7 s(d) 1s