Physics, asked by srishtigunashekar629, 9 months ago

.A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground? After what time will it strike the ground

Answers

Answered by niyati2dinesh
2

Explanation:

Let us assume, the final velocity with which ball will strike the ground be ‘v’ and time it takes to strike the ground be ‘t’

Initial Velocity of ball, u =0

Distance or height of fall, s =20 m

Downward acceleration, a =10 m s-2

As we know, 2as =v2-u2

v2 = 2as+ u2

= 2 x 10 x 20 + 0

= 400

ˆ´ Final velocity of ball, v = 20 ms-1

t = (v-u)/a

ˆ´Time taken by the ball to strike = (20-0)/10

= 20/10

= 2 seconds

Answered by sangeetadas59023
6

Answer:

S,distance=20 m

S,distance=20 ma,acceleration =10m/s^2

S,distance=20 ma,acceleration =10m/s^2v,final velocity=?

S,distance=20 ma,acceleration =10m/s^2v,final velocity=?u,initial velocity=0

S,distance=20 ma,acceleration =10m/s^2v,final velocity=?u,initial velocity=0We know that

S,distance=20 ma,acceleration =10m/s^2v,final velocity=?u,initial velocity=0We know thatv^2-u^2=2aS

S,distance=20 ma,acceleration =10m/s^2v,final velocity=?u,initial velocity=0We know thatv^2-u^2=2aSv^2-0=2×10×20

S,distance=20 ma,acceleration =10m/s^2v,final velocity=?u,initial velocity=0We know thatv^2-u^2=2aSv^2-0=2×10×20400=v^2

S,distance=20 ma,acceleration =10m/s^2v,final velocity=?u,initial velocity=0We know thatv^2-u^2=2aSv^2-0=2×10×20400=v^2v=20m/s

S,distance=20 ma,acceleration =10m/s^2v,final velocity=?u,initial velocity=0We know thatv^2-u^2=2aSv^2-0=2×10×20400=v^2v=20m/st=?

S,distance=20 ma,acceleration =10m/s^2v,final velocity=?u,initial velocity=0We know thatv^2-u^2=2aSv^2-0=2×10×20400=v^2v=20m/st=?We know that 

S,distance=20 ma,acceleration =10m/s^2v,final velocity=?u,initial velocity=0We know thatv^2-u^2=2aSv^2-0=2×10×20400=v^2v=20m/st=?We know that v=u+at

S,distance=20 ma,acceleration =10m/s^2v,final velocity=?u,initial velocity=0We know thatv^2-u^2=2aSv^2-0=2×10×20400=v^2v=20m/st=?We know that v=u+at20=0+10t

S,distance=20 ma,acceleration =10m/s^2v,final velocity=?u,initial velocity=0We know thatv^2-u^2=2aSv^2-0=2×10×20400=v^2v=20m/st=?We know that v=u+at20=0+10tt=2 sec

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