Question

A nall is thrown up vertically returns to the the thrown after 6s. Find 1) the velocity with which it is thrown up 2) the maximum height it reaches, and its position after 4s

Answer 1

$\huge\textbf{ANSWER}$


A) let's consider upward gravity = -9.8 m/s^2
$<b>$

Total time (to and fro)= 6 s, so, upward journey = 6/2 = 3 s

Initial velocity (u) = ?


Final velocity (v) = 0 m/s


From First equation of motion,


Final velocity = Initial velocity +gt


or, 0 = u + (-9.8 X 3)


or, u = 29.4 m/s


So, The velocity at which ball is thrown in the upward direction is 29.4 m/s.


B)

$As \: we \: know \: {v}^{2} - {u}^{2} = 2as \\ \\ putting \: the \: values \: we \: have \\ \\ \\ {0}^{2} - ( {29.4)}^{2} = 2( - 9.8)h \\ \\ h = 44.1m$


C)


$as \: we \: know \: that \: h \: = ut + \frac{1}{2} a {t}^{2} \\ \\ putting \: the \: values \: we \: have \\ \\ 0 + \frac{1}{2} (9.8) \times {1}^{2} \\ \\ = 4.9m$
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Answer 2

_/\_Hello mate__here is your answer--

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The ball takes a total of 6 s for its upward and downward journey. (Time of ascent is equal to the time of descent.)

Hence, it has taken 3 s to reach at the maximum height.

v = 0 m/s

g = −9.8 ms−2

Using equation of motion,

v = u + at

⇒0 = u + (−9.8 × 3)

⇒ u = 9.8 × 3 = 29.4 m/s

Hence, the ball was thrown upwards with a velocity of 29.4 m/s.

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Let the maximum height attained by the ball be h.

u = 29.4 m/s

v = 0 m/s

g = −9.8 ms−2 (upward direction)

Using the equation of motion,

s = ut +1/2 gt^2

⇒ hℎ = 29.4 × 3 −1/2 × 9.8 × 32

⇒ ℎh = 44.1 m

Hence, the maximum height is 44.1 m.

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Ball attains the maximum height after 3 s. After attaining this height, it will start falling downwards.

In this case,

u = 0 m/s

Position of the ball after 4 s − 3 s = 1 s can calculate by Using the equation of motion,

s= ut +1/2gt^2

⇒ h= 0 × 1 +1/2 × 9.8 × 12

⇒ h= 4.9 m

Now, total height = 44.1 m This means, the ball is 39.2 m (44.1 m − 4.9 m) above the ground after 4 seconds.

I hope, this will help you.☺

Thank you______❤

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