An object is dropped from a height H above the ground. .

Question

An object is dropped from a height H above
the ground. This free-falling object requires
0.607 s to travel the last 41 m before hitting
the ground.
From what height H above the ground did
the object fall? The acceleration of gravity is
9.8 m/s
2
.
Answer in units of m.

Answer 1

Travel in last 0.607 sec

h = ut+(1/2)*g*(t^2)

 

41=.607u+0.5*9.8*(0.607^2)

 

solving the above

 

u =64.5 Mt/sec

 

Initial travel---------

Height =(H-41) Mt

Final velocity= 64.5  Mt/sec

Initial velocity=0

 

V^2=U ^2+2*g*h

 

 

64.5^2=0+2*9.8*(H-41)

 

Solving the above equation--

H=253.5 mt