Question

A ball dropeed from a height of 20m if its velocity increases uniformely at the rate of 10ms-2 with what velocity will it strikes the ground after what time will it strike the ground ​

Answer 1

Given:

Height (h) = 20 m

Initial velocity (u) = 0 m/s (Dropped)

Accelration due to gravity (g) = 10 m/s²

To Find:

Final velocity (v) i.e. Velocity with which ball strikes the ground

Time taken to strike the ground (t)

Answer:

$\rm From \: 3^{rd} \: equation \: of \: motion \: we \: have: \\ \boxed{ \bf{ {v}^{2} = {u}^{2} + 2gh}}$

By substituting values in the formula we get:

$\rm \implies {v}^{2} = {0}^{2} + 2 \times 10 \times 20 \\ \\ \rm \implies {v}^{2} = 400 \\ \\ \rm \implies v = \sqrt{400} \\ \\ \rm \implies v = 20 \: m {s}^{ - 1}$

$\therefore$ $\boxed{\mathfrak{Velocity \ with \ which \ ball \ strikes \ the \ ground \ (v) = 20 \ m/s}}$

$\rm From \: 1^{st} \: equation \: of \: motion \: we \: have: \\ \boxed{ \bf{v = u + gt}}$

By substituting values in the formula we get:

$\rm \implies 20 = 0 + 10t \\ \\ \rm \implies 10t = 20 \\ \\ \rm \implies t = \dfrac{2 \cancel{0}}{1 \cancel{0}} \\ \\ \rm \implies t = 2 \: s$

$\therefore$ $\boxed{\mathfrak{Time \ taken \ to \ strike \ the \ ground \ (t) = 2 \ s}}$

Answer 2

Answer:

hello

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’.

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 ball will strike the ground is (v) =  20 ms-1

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

hope it helped you!!!!!!!