Question

a particle is thrown vertically up with initial velocity of 60m/s the distance covered by the particle in first 2 seconds of descent will be

Answer 1

Answer:

Explanation:

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Secondary SchoolPhysics 5+3 pts

3.

A particle is thrown vertically up with initial

velocity of 60 m/s. The distance covered by the

particle in first two seconds of descent will be

(take g = 10 m/s2)

(1) 5 m

(2) 15 m

(3) 20 m

(4) 40 m

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Comments (1) FollowReport by Anchalyadav32 08.06.2019

shamjap

is the answer 40 m

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Answers

The Brain

abhi178

Abhi178 The Sage

answer : option (3) 20m

this question is little different. question want to find distance covered by particle in first two seconds of descent.

particle first moves upward, after reaching a specific height, it falls downward (descending motion). here observing point is that at specific height velocity of particle becomes zero after then it will ready to fall.

so, initial velocity of particle in descending motion, u = 0

distance travelled by particle , s = ut + 1/2 at²

here, u = 0, t = 2s and a = -g = -10m/s²

then, s = 0 + 1/2 (-10m/s²) × (2s)²

= -20m [ here negative sign indicates that particle is moving downward ]

hence, distance covered by particle in first two seconds of descent will be 20m

Answer 2

Answer:

60*2=120

Explanation:

as speed equal to distance by time. and distance equal to speed into time.

given

speed is equal to 60

time is equal to 2

so distance is equal to 60 into 2 is equal to 120