Question

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Answer 1

Answer:

Hey Friend!

Here's your answer,

Given,

Displacement covered in last second = 34.3 m

Using the formula for displacement at the nth second,

we know;

S = U + $\frac{1}{2}$ a ( 2n - 1 )                     [ a = g (10m/s^2) as velocity is along  gravity]

==> 34.3 = 0 + 5 ( 2n - 1 )

==> 10n - 5 = 34.3

==> 10n = 39.3

==> n = 3.93 s

That means at time = 3.93, the ball reaches the ground.

now,

S = Ut + $\frac{1}{2}$ g$t^{2}$

==> S = 0 x t + $\frac{1}{2}$ x 10 x $(3.93)^2$

==> S = 5 x 15.44

==> S = 77.22 m (approx.)

Therefore height of the cliff is 77.22 m

I hope you understand the concept. If you like my answer, please mark it the brainliest and if there is any more confusion be free to ask.

Answer 2

Answer:

refer to the pic above for ur answer

Explanation:

hope it helps you ^_^