Question

Date
A ball is gently dropped from a height of
20m. If its velocity increases uniformly
at the rate of 10 m s-2 , with what
velocity will it strike the
ground. After what time will it strike
the ground?
​

Answer 1

Explanation:

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Answer 2

Given, Initial velocity, u = 0 (as it is dropped)

Height, h = 20m.

Acceleration, a = ${10}ms^{2}$

Final velocity, v = ?

Time, t = ?

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

$\textsf{Here}$, S = h

${v}^{2} = {u}^{2} + 2as \\ = > 0 + 2 \times 10 \times 20 \\ 0 + 400 \\ \\ v = \sqrt{400} \\ v = {20}ms^{1}$

Time = ?

$\textsf{Using\:1st\:equation\:of\:motion:-}$

V = u + at

$\frac{v - u}{a} = t \\ = > \frac{20 - 0}{10} = \frac{20}{10} \\ = > 2s$

Therefore, time = 2s.