Question

A hammer is dropped and it falls for 9 seconds 9 seconds before hitting the ground. Determine how far it falls, assuming an acceleration due to gravity of − 9.8 m/s 2 −9.8 m/s2 and no wind resistance.

Answer 1

Answer:

396.9 meter

Step-by-step explanation:

If a represents the acceleration,

Then according to the question,

a = -9.8 m/s²,

$\frac{dv}{dt}=-9.8$     ( acceleration = change in velocity with respect to time ),

$dv = -9.8dt$

Integrating both sides,

$v=-9.8t + C$

When t = 0 seconds, v = 0 m/s,

$0 = -9.8(0) + C$

$\implies C = 0$

$\implies v=-9.8t$

$\frac{dx}{dt}=-9.8t$ ( velocity = change in position with respect to time ),

$dx = -9.8tdt$

Integrating again,

$x=-4.9t^2 + C'$

When t = 0 , x = 0 meters,

$0=-4.9(0) + C'\implies C'=0$

Hence, the final equation that shows the position of the hammer after t seconds,

$x=-4.9t^2$

If t = 9 seconds,

$x = -4.9(9)^2 = -4.9(81) = -396.9$  ( negative sign shows the fall ),

Therefore, it will fall 396.9 meter in 9 seconds