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.
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.
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]
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]
