Question

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Answer 1

Answer:

Given :-

Ball returns to the ground after 4 seconds.

∴ The time taken by the ball to reach maximum height is 2s.

i)

Final velocity, v = 0 m/s

Time taken, t = 2s

Acceleration due to gravity, g = -10 m/s

\boxed{\bf v = u+gt}

v=u+gt

: \implies \sf 0 = u+ -10 \times 2:⟹0=u+−10×2

: \implies \sf \bf u = 20 m/s:⟹u=20m/s

ii)

\boxed{\bf h= ut+\dfrac{1}{2}gt^2}

h=ut+

2

1

gt

2

: \implies \sf h = 20 \times 2+ \dfrac{1}{2} \times -10 \times 2 \times 2:⟹h=20×2+

2

1

×−10×2×2

: \implies \sf h = 40-20:⟹h=40−20

: \implies \bf h = 20m:⟹h=20m

iii)

The ball takes 2s to reach the maximum height.

After 3s, it starts to fall down.

Time taken for the downward journey = 4-3 = 1s

: \implies \sf h = 0+ \dfrac{1}{2} \times 10 \times 1^2:⟹h=0+

2

1

×10×1

2

: \implies \sf h = 5m:⟹h=5m

Position after 3s = 20-5

= 15m

Answer 2

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