Question

A stone is dropped from the top of a tower 44.1m high. (i) How long does it take to reach the ground? (ii) With what speed does it hit the ground?

Answer 1

Initial speed of the stone (u) = 0 m/s

Let the final speed be v.

Height of the tower (H) = 44.1 m

Acceleration due to gravity (g) = 9.8 m/s^2

Now, using the seconds equation of motion:

$H = ut + \frac{1}{2}gt^2$

$44.1 = 0 + \frac{1}{2}\times 9.8 \times t^2$

$t = \sqrt{\frac{44.1}{4.9} }$

(I) t = 3 seconds

(II) v = u + at

v = 0 + 9.8 \times 3

v = 29.4 m/s

Hence, the final speed is 29.4 m/s

Answer 2

Answer:

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