Question

$Q)..A \: ball \: is \: gently \: dropped \: from \: a \: height \: of \: 20m. \\ If \: its \: velocity \: increases \: uniformly \: at \: 10m \: {s}^{ - 2} , \\ with \: what \: velocity \: will \: it \: stricks \: the \: ground \: ? \\ After \: what \: time \: will \: it \: strike \: the \: ground?$
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Answer 1

Answer:

$v = 20 \: {ms}^{ - 1} \: and \: t = 2s.$

Explanation:

Given that:

• Initial velocity,u = 0 ms·¹
• Acceleration,a = 10·²
• Height,h = 20m

To find:

• Final velocity,v = ?
• Time,t = ?

Formula to know:

• Equations of motion
• v = u+at
• s = ut+ ½at²
• v² = u² = 2as✔️

Solution:

Using the formula v²-u² = 2as

=> v²-0 = 2×10×20

=> v² = 400

=> v = √400

$= > v = 20 \: m {s}^{ - 1}$

Again using the formula v = u+at

$20 = 0 + 10 \times t \\ t = \frac{20}{10} = 2s \\ hence \: v = 20 {ms}^{ - 1} \: and \: t = 2s \: \: ans.$

Answer 2

Solution:

Using the formula v²-u² = 2as

=> v²-0 = 2×10×20

=> v² = 400

=> v = √400

=>v=20ms ^−1

Again using the formula v = u+at

20=0+10×t

t= 10/20

=2shencev=20ms −1 andt=2sans.