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lubasha [3.4K]
2 years ago
6

A ball is gently dropped from a height of 20 m. If its velocity Increases uniformly at the rate of 10 m s2, with what velocity w

ill it strike the ground? After what time will it strike the ground?​
Physics
2 answers:
Pie2 years ago
5 0

Answer:

20 m/s

Explanation:

Height (s)= 20m

acceleration (a) = 10 m/s^2

v=?

we know,

V^2= U^2 + 2as. (U is initial velocity)

or,. V^2 = 0 + 2 × 10 × 20

or, V^2 = 400

or,. V = 20 m/s

belka [17]2 years ago
5 0

Explanation:

Initial Velocity of the 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

v2 = (2 x 10 x 20 ) + 0

v2 = 400

Final velocity of ball (v) = 20 ms-1

t = (v – u)/a

Time taken by the ball to strike (t) = (20 – 0)/10

t = 20/10

t = 2 seconds

The final velocity with which the ball will strike the ground is (v) = 20 ms-1

The time it takes to strike the ground (t) = 2 seconds

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B)

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F=qvB

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F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

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r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

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T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

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