If a body of mass 3 kg is dropped from the top of a tower of height 25 m , then kinetic energy after 3 sec will be:

Question

If a body of mass 3 kg is dropped from the top of a tower of height 25 m , then kinetic energy after 3 sec will be: (A)  557 textJ  (B)  246 textJ  (C)  1048 textJ  (D)  1297 textJ

Tardigrade Admin · Accepted Answer

Here: mass m=3 kg, height h=25 m and time t=3 sec Final velocity of body after 3 sec, v=u+g t=0+9.8 × 3=29.4 m / s Hence, kinetic energy of the body is given by =(1/2) m v2=(1/2) × 3 ×(29.4)2 ≈ 1297 J

Q. If a body of mass $3 k g$ is dropped from the top of a tower of height $25 m$ , then kinetic energy after $3 sec$ will be:

$557 J$

$246 J$

$1048 J$

$1297 J$

Here $:$ mass $m = 3 k g$ , height $h = 25 m$ and time $t = 3 sec$
Final velocity of body after $3 sec$ ,
$v = u + g t = 0 + 9.8 \times 3 = 29.4 m / s$
Hence, kinetic energy of the body is given by
$= \frac{1}{2} m v^{2} = \frac{1}{2} \times 3 \times (29.4)^{2}$
$\approx 1297 J$

Q. If a body of mass 3kg is dropped from the top of a tower of height 25m , then kinetic energy after 3sec will be:

557J

246J

1048J

1297J