Question

A ball is gently dropped from a height of 20m if it velocity increases uniformly at the rate of 10m/s² with what velocity will it strike ground After what time will it strike the ground ​

Answer 1

Answer:

20m/s

Explanation:

H=20m

Acceleration due to gravity=10m/s^2

v=√(2gH)

=√(2×10×20)

=√400

=20m/s

Answer 2

Aɴꜱᴡᴇʀ

V = 20 m/s

T= 2 sec

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Gɪᴠᴇɴ

✭Initial velocity (u) = 0

✭Height of fall ,s= 20 m

✭Acceleration, a = 10 m/s²

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ᴛᴏ ꜰɪɴᴅ

➠ Time of fall, t = ?

➠Final velocity, v = ?

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Sᴛᴇᴘꜱ

As we know

$\mapsto \tt{v}^{2} - {u}^{2} = 2as$

$\tt \leadsto{v}^{2} - 0 = 2 \times 10 \times 20 \\ \\ \tt \leadsto {v}^{2} = 400 \\ \\ \tt \leadsto v = \sqrt{400} \\ \\ \tt \green{ \leadsto{}v = 20 \: m/s}$

☞ Now for t we use the formula v = u +at

$\tt \dashrightarrow{}20 = 0 + 10 \times t \\ \\ \tt \dashrightarrow \frac{20}{10} = t \\ \\ \tt \pink{ \dashrightarrow{}t = 2s}$

Hence, time of fall is 2 sec

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