Question

A stone is dropped from a high tower and 9.21 seconds later it hits the ground. How high is the tower?

Answer 1

Answer:

S = 415.64m

Explanation:

Given the following data;

Time, t = 9.21secs

Since it's a free fall, acceleration due to gravity = 9.8m/s²

Initial velocity, u = 0

To find the height of the tower, we would use the second equation of motion;

$S = ut + \frac {1}{2}at^{2}$

Where;

• S represents the displacement or height measured in meters.

• u represents the initial velocity measured in meters per seconds.

• t represents the time measured in seconds.

• a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

$S = 0*9.21 + \frac {1}{2}*(9.8)*9.21^{2}$

$S = 0 + 4.9*84.8241$

$S = 4.9*84.8241$

Height, S = 415.64m

Therefore, the tower is 415.64m in height.