A stone is dropped from the top of a tower.If it hits the ground after 10 seconds, what is the height of the tower?
We can use the formula S=ut+([math]1/2)[/math]a[math]t^2[/math]Where u is initial velocity, ‘s’ is distance travelled (Height of the tower in this case), ‘t’ is time travelled and ‘a’ is the constant acceleration.So , here as the object is dropped , the initial velocity ,u is 0Acceleration is the acceleration due to gravity ‘g’ which is constant and has value of 9.8m/[math]s^2[/math]Time travelled is given as t=10 secondsSubstituting the values in the equation s=ut+([math]1/2)[/math]a[math]t^2 [/math]we gets=0 * 10 + ([math]1/2)[/math]* 9.8 * 10 *10s=490 metersSo, the height of the tower is 490 meters