Question

A ball is gently dropped from a height of 20m if its velocity increases uniformly at the rate of 10ms-2 , with what velocity will it strike the ground​

Answer 1

Answer:

Initial velocity, u = 0 (dropped gently); Acceleration, a = 10 m/s2

Height, s = 20 m; Final velocity, v =? And Time taken, t =?

Use third equation of motion:

begin mathsize 12px style straight v squared equals straight u squared plus 2 as

rightwards arrow straight v squared equals 0 plus 2 cross times 10 cross times 20

rightwards arrow straight v squared equals 400

rightwards arrow straight v equals 20 space straight m divided by straight s

Using space the space second space equation space of space motion comma space we space get

straight s equals ut plus 1 half at squared

rightwards arrow 20 equals 0 cross times straight t plus 1 half cross times 10 cross times straight t squared

rightwards arrow straight t squared equals 4

rightwards arrow straight t equals 2 space straight s end style

Explanation:

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Answer 2

Explanation:

Data:

h=20m

vi=0m/s

g= 10ms-2

Vf=?

Solution:

Using 3rd equation of gravity:

2gh=$vf^{2}$-$vi^{2}$

2* 10* 20= $vf^{2}$- $0^{2}$

400=    $vf^{2}$

$\sqrt{400}=\sqrt{vf^{2} } \\20 m/s=vf$

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