Question

A ball thrown up vertically returns to the thrower after 6 s. Find (a) the velocity with which it was thrown up. (b) the maximum height it reaches, and (c) its position after 4s.

Answer 1

Intial Velocity(u)=0 m/s

Time(t)=6 seconds

Accerlation due to gravity=10 m/s
As against gravity (a)=-10 m/s
Final velocity (v)=??

Using Netwon's law
v=u+at
v=0+10×6
v=60 m/s

(A)=60 m/s

(B) Distance(s)=??

$s = ( \frac{v + u}{2} ) \times time \\ s = ( \frac{60 + 0}{2} ) \times 6 \\ s = 30 \times 6 \\ s = 180 \: metres \\ \\ maximum \: distance \: it \: reach \\ = 180 \: metres$
C.In 6 it will complete=180 METRES
In 1 sec =180/6
1 sec=30 meter

So in 4 second =4×30
=120 METRES

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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