Question

A ball is gently dropped from a height of 20m. If its velocity increases uniform,ly at the rate of 10ms^-2, with what velocity will it stike the ground? After what time will it stirke the ground?

Answer 1

Answer:

➪ Velocity of ball of hitting ground

= 20 m/s

➪ Time taken ball to reach ground

= 2 seconds

Step by step explanations :

Given that,

A ball is gently dropped from a height of 20m

here,

initial velocity of the ball = 0 m/s

[ball was initially at rest]

height from which ball is dropped = 20 m

also given that,

its velocity increases uniformly at the rate of 10m/s²

so,

here,

acceleration of the ball = 10 m/s²

let the velocity of the ball when it will hit the ground be v

now,

we have,

initial velocity(u) = 0 m/s

final velocity(v) = v

acceleration(g) = 10 m/s²

height(h) = 20 m

by the gravitational equation of motion,

v² = u² + 2gh

putting the values,

v² = 0² + 2(10)(20)

v² = 400

v = 20 m/s

so,

Velocity of ball of hitting ground

= 20 m/s

now,

let the time taken by the ball to reach ground be t

also

v = u + gt

again putting the values,

20 = 0 + 10t

10t = 20

t = 20/10

t = 2 s

so,

Time taken ball to reach ground

= 2 seconds

Answer 2

$\bold {\huge {Your ~answer :-}}$

$\bf\huge\boxed{20m/s,\:2sec}$

--------------------------------------------------

Given & to find —

initial velocity of ball,u = 0

Final velocity of ball, v= ?

Distance through which the balls falls,s = 20 m

Acceleration, a = 10 m/s²

Time of fall, t =?

--------------------------------------------------

We know

${v}^{2} - {u}^{2} = 2as$

$= > {v}^{2} - 0 = 2 \times 10 \times 20$

=> v = 20 m/s

Therefore velocity while strike the ground is 20 m/s

--------------------------------------------------

Now for t we know

v = u + at

=> 20 = 0 + 10 t

=> t = 20/10

=> t = 2 sec

Therefore, time of fall is 2 sec

$\huge{\red{\ddot{\smile}}}$