A ball of mass 0.5 kg is dropped from the height of 10 m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is ....... m. (Use g=10 m / s 2).

Question

A ball of mass 0.5 kg is dropped from the height of 10 m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is ....... m. (Use g=10 m / s 2). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

v2=u2+2 a s 100=0+2(10) s S=5 m Height from ground =10-5=5 m

Q. A ball of mass $0.5 k g$ is dropped from the height of $10 m$ . The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is ....... $m$ . (Use $g = 10 m / s^{2}$ ).

$v^{2} = u^{2} + 2 a s$
$100 = 0 + 2 (10) s$
$S = 5 m$
Height from ground $= 10 - 5 = 5 m$

Q. A ball of mass 0.5kg is dropped from the height of 10m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is ....... m. (Use g=10m/s2).

v2=u2+2as 100=0+2(10)s S=5m Height from ground =10−5=5m

Q. A ball of mass $0.5 k g$ is dropped from the height of $10 m$ . The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is ....... $m$ . (Use $g = 10 m / s^{2}$ ).

$v^{2} = u^{2} + 2 a s$
$100 = 0 + 2 (10) s$
$S = 5 m$
Height from ground $= 10 - 5 = 5 m$