Question

Q3) A ball thrown up vertically returns to the thrower after 8 s. Using g = 10 m/s2 , find a) the velocity with which it was thrown up, b) the maximum height attained by the ball?​

Answer 1

Answer:

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,

v=u+at

∴0=u+(−10)×3

⟹u=30 m/s

• (b). The maximum height reached by the ball

$h=ut+21at2 \\ </p><p></p><p>h=30×3+21(−10)×32   \\            </p><p></p><p>h=45 m  </p><p></p><p>$

• (c). After 3 second, it starts to fall down.

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

$d=0+21at′2                        where   t′=1 s \\ </p><p></p><p>d=21×10×(1)2    =5 m  </p><p></p><p>$

∴ Its height above the ground, h

′

=45−5=40 m

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

HOPE IT HELPS YOU (≡^∇^≡)