Question

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A ball thrown up vertically returns to the thrower after 6 s.${\pmb{\textsf{Find \; : }}}$

(a) the velocity with which it was thrown up .
(b) the maximum height it reaches .
(c) its position after 4 s .


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Answer 1

The ball returns to the ground after 6 seconds.

Thus the time taken by the ball to reach to the maximum height

(h) is 3 seconds i.e t=3 s

Let the velocity with which it is thrown up be u

(A) For upward Motion

$\green{\boxed{\bf \: v = u \: + at}}$

$\tt0 = u + ( - 10) \times 3$

$\sf⟹ \: u \: = 30 \: m/s$

The Maximum height reached by ball :-

$\leadsto \tt \: h = ut + \frac{1}{2} \: {at} \: ^{2}$

$\bf \: h = 30 \times 3 + \frac{1}{2}( - 10) \times {3}^{2}$

After three second it will start to fall down .

Let the distance by which it fall in 1 s be d

$\sf \: d = 0 + \frac{1}{2} \: \: {at} \: \: ^{2}$

$\bf\boxed{\green{ {t} \pink{ \: = 1s}}}$

Then :-

$\bf \: d = \frac{1}{2} \times 10 \times (1) {}^{2}$

∴ Its height above the ground,

$\red{\bf \: h = 45 - 5 = 40 cm}$

Hence after 4 s, the ball is at a height of 40 m above the ground

( So (c) one is correct )