Question

$\red{\large\mathfrak {Solve \: its \: very \: urgent}}$
A ball is thrown downwards from a height of 20 m. If its velocity increases uniformly at the rate of 10m/s². With what velocity will it strike the ground. ​

Answer 1

Answer

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

Given parameters

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

Answer 2

Explanation:

Given, initial velocity of ball, u=0

Final velocity of ball, v=?

Distance through which the balls falls, s=20m

Acceleration a=10ms−2

Time of fall, t=?

We know

v²−u²=2as

or v²−0=2×10×20=400 or v=20ms−1

Now using v=u+at we have

20=0+10×t or t=2s

=20m/s,2s