Question

6. A ball thrown vertically upwards takes 4 s to retu
to the ground. Calculate the following:
(a) maximum height reached
(6) velocity with which it was thrown upward.
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Answer 1

Answer:

Explanation:

Given :-

Time taken = 4 sec

Final velocity = 0  m/s

Acceleration = -9.8 m/s²

To Find :-

(a) Maximum height reached

(b) Velocity with which it was thrown upward.

Formula to be used :-

1st Equation of Motion i.e v = u + at

2nd Equation of Motion i.e v² = u² + 2as.

Solution :-

At First we will solve (b) then (a)

(b) Velocity with which it was thrown upward.

⇒ v = u + at

⇒  0 = u + (-9.8) (4)

⇒  0 = u - 39.2

⇒  39.2 m/s = u

(a) Maximum height reached

⇒ 2as = v² - u^2

⇒  2 (-9.8) (s) = 0 - (39.2)²

⇒ 19.6 × s = 1536.64

⇒  s = 1536.64/19.6

⇒ s = 78.4 m

Hence, the maximum height reached is 78.4 m and the velocity with which it was thrown upward 78.4 m.

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Answer 2

Given

A ball is thrown vertically upwards and the time take taken to return is 4 sec

To find

$\bullet \sf maximum \:height\: reached\: by\: the \:body \\ \bullet \sf Velocity\: with \:which \:it \:is \:thrown\: upwards$

Explanation:

when the body is thrown upwards at its maximum height the velocity attained by the body is 0

$\sf \:let \: initial \: velocity \: with \: which \: the \: body \: is \ \\ \sf \: thrown = u$

$\sf \: final \: velocity \: (v) = 0$

$\sf \: acceleration = - g = - 9.8m {s}^{ - 1}$

$\sf \: time = 4sec$

From the formula

$\underline {\sf v=u+at}$

$0 = u + ( - 9.8)(4)$

$= > u = 39.2 m{s}^{ - 1}$

$\boxed {u = 39.2m {s}^{ - 1} }$

So the velocity with which it was thrown is 39.2 m/s

$\sf \: let \: the \: maximum \: height \: = s$

$\sf \: time \: = 4s$

$\sf \: u = 39.2 m{s}^{ - 1}$

$\sf \:v = 0$

$\sf a = - 9.8 m{s}^{-1}$

applying

$\underline{ \sf \: {v}^{2} - {u}^{2} = 2as}$

${0}^{2} - {(39.2)}^{2} = 2( - 9.8)s$

$s = \dfrac{39.2 \times 39.2}{19.6}$

$= > s = 39.2 \times 2 = 78.4m$

$\boxed{s = 78.4m}$

The maximum height attained by the body is 78.4m