Question

Please help me with thos question. I am not able to solve it.
Please do not write irrelevant answers. I will mark you as brainliest.



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Answer 1

Answer:

s = ut + ½at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, we know u = 0, a = g = 9.81m/s^2, t = 4s and we want to find the distance, s, so we use equation (1)

acceleration (this is a lesson – suggest you commit these to memory):

s = ut + ½at^2

s = 0 + ½(9.81)(4^2)

s = 78.48m

The tower is 78.48m high.

Answer 2

Explanation:

buddy I hope it is usefull to you