Question

A stone is dropped from the top of a tower if its travel 34.3 m in the last second before its reaches the ground ,find the height of the tower (g=9.8m/s2)​

Answer 1

The Given Information is:

t in the last second = 1 s

s in the last second = 34.3 m

u = 0 m/s. (Since the stone falls from the cliff with no velocity)

a=g = 9.8 m/s²

Now,

v = s/t

v = 34.3m/1s

v = 34.3m/s

But this is the average velocity in the last second.

With the the acceleration of 9.8 m/s², we can calculate the final velocity.

V²-U²=2as

(V+U)(V-U)=2as

We know, v=u+at

Therefore v-u =at

Substituting in previous equation.

(V+U)(at)=2as

(V+U)=68.6m/s (After substituting the values)

Solving for V, we get V=39.2m/s

Now according to Newton's laws of motion,

V²-u²= 2as

(39.2m/s)²-(0)²= 2(9.8)(S)

(1536.64)/19.6= S

S= 78.4 m

Therefore the cliff is 78.4 metres high

Note:

>v-u = at = 9.8 use this in solving for V

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