Question

A stone dropped from the top of tree covers 6 metre height in last 0.2 seconds of fall .find height of the tree

Answer 1

Answer:

A stone is drop from a tree takes 0.2s to cross the last 6m before hitting the ground. What is the height of tree?

Let:

Tree height =h

Initial velocity =u

Current velocity =v

Acceleration (due to gravity) =g

Time =t

The first thing to note will be that to get the tree height

h=12gt2

We will need the time it takes for the stone to get to the ground i.e. how long it falls for. To do this, we will use the information given to us about the final 6m. The 6m travelled can be expressed as.

6=ut+12gt2

=u(0.25)+12g(0.25)2

⟹u=24−18g

So how does this help us? Well now we can use another equation to find the time taken for the first section of the fall. Note

v=u+gt

So let the current velocity be 24−18g as above. We know the ball was dropped, so initial velocity is zero, which all in all gives:

24−18g=gt

⟹t=24g−18

So time taken for first portion of the fall, t1 , is

24g−18

The second portion, t2 , was given to be 0.25 (1/4) seconds, adding these gives us the full length of the trip:

t=t1+t2=24g−18+14

=24g+18

Now back to the first equation! Simply substitute these values in to get

h=g2(24g+18)2

Now plug in whatever value you’re using for g and you’re done!

Hope this all makes sense now :)