Question

A ball thrown up vertically returns to the thrower after 6s
Find
a. The velocity with which it was thrown up.
B.The maximum height it reaches
C.The position after 4s​

Answer 1

$\blue{\bold{\underline{\underline{Consideration:-}}}}$

• Time taken taken is 6sec
• Half Time is 3sec
• Final velocity is 0 m/s
• acceleration is gravity which is 10m/s²
• acceleration when downward is -10m/s²

$\red{\bold{\underline{\underline{To\:Find:-}}}}$

• The velocity with which it was thrown up
• The maximum height it reaches
• The position after 4sec

$\green{\bold{\underline{\underline{Solution:-}}}}$

➣Case - 1

$⟹v = u + at \\ \\ ⟹ 0 = u + ( - 10)(3) \\ \\⟹ 0 = u - 10 \times 3 \\ \\⟹ 0 = u - 30 \\ \\⟹ 30 = u$

$\pink{\bold{\underline{\underline{The\: velocity\:with \:which\:it\:was \: thrown\:up\:is\: 30m/s}}}}$

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➢Case - 2

$by \: using \: 2nd \: equation \: of \: motion \\ \\⟹ s = ut + \frac{1}{2} a {t}^{2} \\ \\ ⟹s = 30(3) + \frac{1}{2} (10) {(3)}^{2} \\ \\ ⟹s = 45m$

$\purple{\bold{\underline{\underline{The\: maximum\: height\: it \: reaches\: is\: 45m}}}}$

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➢Case - 3

$total \: time \: is \: 6s \\ \\ by \: using \: 2nd \: equation \: of \: motion \\ \\ ⟹s = ut + \frac{1}{2} a {t}^{2} \\ \\ ⟹s = 0 + \frac{1}{2} (10) \\ \\ ⟹s = 5m \\ \\⟹ 45m - 5m = 40m$

$\orange{\bold{\underline{\underline{Position\:after\:4sec\: is\: 40m}}}}$

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$\huge\underline\mathfrak\purple{Thanks}$

Answer 2

Answer:

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Explanation:

Time to reach Maximum height ,

t = 6/2 = 3 s.

v = 0 (at the maximum height )

a = - 9.8 m s-²

◆ a) Using, v = u + at, we get

0 = u - 9.8 × 3

or, u = 29.4 ms-¹

Thus, the velocity with which it was thrown up = 29.4ms-¹

◆ b) Using, 2aS = v² - u², we get

S = v²- u²/2a

= 0 - 29.4 × 29.4/- 2× 9.8

= 44.1 m

Thus, Maximum height it reaches = 44.1 m.

◆ c) Here, t = 4s. In 3 s, the ball reaches the maximum height and in 1 s it falls from the top.

Distance covered in 1 s from maximum height,

S = ut + 1/2at ²

= 0 + 1/2 × 9.8 × 1

= 4.9 m

Therefore, The ball will be 4.9 m below the top of the tower after 4 s.

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