Question

A ball thrown up vertically returns to the thrower after 12s. Find:velocity maximum height it reaches position after 4 seconds fast im in class

Answer 1

Answer:

t = 6/2 = 3 s.

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Answer 2

(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+

2

1

at

2

h=30×3+

2

1

(−10)×3

2

h=45 m

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

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

d=0+

2

1

at

′2

where t

′

=1 s

d=

2

1

×10×(1)

2

=5 m

∴ 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.