Question

If an object is dropped from the top of a building and it reaches the ground at t = 4 s , then the height of the building is (ignoring air resistance) (g = 9.8 ms⁻²)
(a) 77.3 m
(b) 78.4 m
(c) 80.5 m
(d) 79.2 m

Answer 1

An object is thrown from the top of a building.

So, the initial velocity of the object is zero.

Also, we are given that the time taken to reach the ground is 4 seconds. And there's the gravitational acceleraton $g=9.8 \, \, m/s^2$ acting in the downward direction.
Our data is:

Initial Velocity = u = 0
Time Taken = t = 4 s
Acceleration = $g = 9.8 \, \, m/s^2$
Height of Tower = h (say)

We are assuming the height of tower to be $h$

Now, we will use one of the equations of motion:


$s =ut + \frac{1}{2}at^2 \\ \\ \implies h = ut + \frac{1}{2}gt^2 \\ \\ \implies h = (0)(4) + \frac{1}{2} (9.8) \times (4)^2 \\ \\ \implies h = 0 + \frac{9.8\times 16}{2} \\ \\ \implies h = 9.8 \times 8 \\ \\ \implies \boxed{h = 78.4 \, \, m}$

Thus, the Height of the Tower is 78.4 metres


The answer is Option (b) 78.4 m