Question

A body dropped from a tower covers a distance 45 m in last second of its motion . Find the height of the tower ( g = 10 m / s2 )​

Answer 1

Given:-

Initial velocity(u)=0

Acceleration due to gravity(g)=10m/s²

Distance travelled in the last 1 second=45m

To find:-

Height of the tower(h)

Solution:-

Let the total distance travelled be h

Time taken to travel (h-45)=t-1 second

By h=1/2gt²-----

=>h=1/2×10×t²

=>h=5t² -------(1)

Now-----

=>(h-45)=1/2×g×(t-1)²

=>5t²-45=1/2×10(t-1)² (Putting value of h from eq. 1)

=>5t²-45=5(t-1)²

=>5t²-45=5t²+5-10t

=>-45-5= -10t

=>-50= -10t

=>t=-50/-10

=>t=5s

Now,by putting value of t in equation 1-----

=>h=5(5)²

=>h=125

Thus,height of the tower is 125m.

Answer 2

Answer:

125

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