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Question

# 1.A stone falls from a building and reaches the ground in 2.5 seconds latter . How high is the building? 2. A stone dropped from a height of 20meter i) How long it will take to reach the ground. ii)what will be it's speed when it hits the ground.

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Solution

## 1. We can use the formula S=ut+(1/2)at2 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 0 Acceleration is the acceleration due to gravity ‘g’ which is constant and has value of 9.8m/s2 Time travelled is given as t=2.5 seconds Substituting the values in the equation s=ut+(1/2)at2we get s=0 * 2.5 + (1/2)* 9.8 * 2.5^2 s=30.625 meters So, the height of the tower is 30.625 meters 2. H=20m/s u=0 v=? t=? v²-u²=2gh v²-o²=2*10m/s * 20m v=√400 v=20m/s[velocity with which stone hit the ground] v=u+gt 20=0+10*t t=2s[time to reach the ground]

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