Question

A ball is dropped from the top of the tower of height h. The total distance cover by the ball in the last 5 second of its motion is equal to the distance covered by its in the first three seconds. What is h?

Answer 1

we know that

s=ut+1/2 at²

now for the distance travelled in first 3 seconds

u=0,a=9.8 m/ s²,t=3 sec

so S =9.8*3*3/2=44.1 m

now distance travelled in last 2 seconds

u=u,t=5

S 1 =5u +9.8*5*5/2

S 1=5 u +122.5

since given distance covered in both case is same

S=S 1

44.1=5 u+122.5

5 u=-78.4

u= -15.68 m/s

so final velocity V=u+at

V =-15.68+9.8*5

V =-15.68+49

V =33.32 m /s

as V²-u²=2ah

1110.22-245.86=2*9.8*h

864.36=19.6*h

h=864.36/19.6=44.1

so height of tower is 44.1 m