Question

A ball is gently dropped from a height of 20m if its velocity in crease uniformly at the rate of 10m/s2,with what velocity will it strike the ground?after what time will it strike the ground ?

Answer 1

Answer:

It will strike the ground with the velocity of 20m/s and it will strike the ground after 2 seconds.

Explanation:

Given-

Acceleration (a) = 10 m/s²

Height (s) = 20 m

Initial Velocity (u) = 0 m/s

Final Velocity (v) = ?

Time of impact (t) = ?

First let's calculate the Final Velocity (v):

From 3rd equation of motion- 2as = (v)²- (u)²

=> 2(10)(20) = (v)² - (0)²

=> 400 m/s = v²

=> √400 m/s = v

=> 20 m/s = v

Now we can calculate the Time of impact (t):

From 1st equation of motion- v = u + at

=> 20 = 0 + (10)t

=> 20 = 10t

=> 20/10 = t

=> 2 seconds = t

Hope you got your answer. Kindly mark it as brainliest !

Answer 2

Answer:

Final velocity = 20 m/s

Time taken =2 secs

Explanation:

Given Information -

• Initial velocity (u) = 0m/s
• Acceleration (a) =10m/s²
• Height/distance (s) =20 m

ㅤ

To Find -

• Final velocity (v)
• Time (t)

ㅤ

Solution -

$\underbrace\mathscr\red{Finding\:Final\:Velocity(v)-}$

Formula - $\underline{\boxed{\sf\purple{2as = v² -u² }}}$

ㅤ

Answer -

$2(10 \times 20) = v{}^{2} - {0}^{2} \\ \\ 400 = {v}^{2} \\ \\ \sqrt{400} = v \\ \\ 20 = v$

Final velocity is 20m/s.

ㅤ

$\underbrace\mathscr\red{Finding\:Time(t)-}$

Formula - $\underline{\boxed{\sf\purple{v = u+at}}}$

ㅤ

Answer-

$20 = 0 + 10 \times t \\ \\ \frac{20}{10} = t \\ \\ 2 = t$

Time here is 2 secs.