Question

A ball is dropped off the top of a building. It accelerated downward at 9.8m/s2 and it hits the ground 3.9 seconds later. Neglecting air resistance, how high is the cliff?

Answer 1

Answer: building's height would be 74.61m

Explanation:

Basic kinematics, negating drag and assuming ideal conditions, we use the equation:

$h=\frac{1}{2} gt^{2}$

g= $9.81 m/s^2$

t=$3.9^{2}$=15.21

input into the equation below.

h=$\frac{1}{2} (9.81)(25)$= 74.61m this would be the height of the tower.

Answer 2

Answer:

Given;

t = 3.9s

g = 9.8m/s²

Required;

Height, S.

from 3rd eqn of motion,

- S = ut - gt²/2

u = 0m/s since motion started at rest.

- S = - 9.8*(3.9) ²

2

S = 74.53m

The cliff is 74.53m high.