Question

An apple is dropped from a 260m high building. How long will it take to hit the ground in
seconds)?

Answer 1

Answer:

7.28 s

Explanation:

s=-260

u=0

assume a=-9.81

t=?

s=1/2at^2 + ut

-260=1/2 (-9.81) t^2 +0t

t=7.28 s

Answer 2

The time taken for the apple to hit the ground from the high building is 7.28s.

Given the data in the question;

Before the apple was dropped, it was initially at rest, so

• Initial velocity; $u = 0 m/s$

• Distance or height from which it was dropped from; $s = 260m$

 

To find the time it will take for the apple to hit the ground

We use the Second Equation of Motion:

$s = u^2 + \frac{1}{2}at^2$

Where s is the distance or height from which it was dropped, u is its initial velocity, t is time and a is the acceleration due to gravity.

We know that earth's gravitational field or acceleration due to gravity; $g = 9.8m/s^2$

Since the apple is falling, it under gravity; $a = g = 9.8m/s^2$

We substitute our values into the equation and find t

$260m = (0m/s)^2 + (\frac{1}{2} \ *\ 9.8m/s^2\ *\ t^2 )\\\\260m = 4.9m/s^2 \ * \ t^2\\\\t^2 = \frac{260m}{4.9m/s^2}\\\\t^2 = 53.06^2\\\\t = \sqrt{53.06s^2} \\\\t = 7.28s$

Therefore, the time taken for the apple to hit the ground from the high building is 7.28s

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