Question

Two spheres of 1 kg and 5 kg, are dropped
simultaneously from the same height. Which one will
reach the ground first ?​

Answer 1

Answer:

Heya mate,

Both will reach simultaneously, if air friction is negligible.

Hope it helps you.

Answer 2

Hey mate

• if air friction is negligible then both will reach simultaneously on the ground

BUT

• If air friction is not neglible then

Stokes force comes into action and final velocity (with which body reaches the ground) is given by

$v = \frac{2}{9 \: \: \: \: \: \: \: \eta} r {}^{2} g( \rho b - \rho m) \\ where \: \\ v = terminal \: velocity(with \: which \: body \: reaches \: the \: ground) \\ r = \: radius \: of \: body \\ g \: = acceeration \: due \: to \: gravity \\ \rho b \: = density \: of \: body \\ \rho m = density \: of \: medium \\ \eta = coefficent \: of \: viscousity$

• sphere 1

$mass \: = volume \times \rho b$

where density for both bodies is same

$1kg = (\frac{4}{3} \pi r {}^{3} ) \rho b$

where r is radius of 1kg body

$\frac{3}{4\pi} = r {}^{3} \times \rho b$

• sphere 2

$5 = \frac{4}{3} \pi d {}^{3} \rho b$

where d is radius of 5kg body

$from \: terminal \: velocity \: expression \: \\ it \: is \: clear \: that \\ v \infty (radius) {}^{2}$

the body qith greater radius will reach first

$1kg \: body \\ r {}^{3} = \frac{3}{4\pi \rho b}$

$d {}^{3} = \frac{(5)3}{4\pi \rho b}$

$hence \: d > r$

i.e. radius of 5kg body is greater than 1kg body

hence v of 5kg body > 1kg body

HENCE IF AIR FRICTION IS TAKEN IN ACCOUNT HEAVIER BODY WILL REACH FIRST

hence