In a shotput event an athlete throws the shotput of mass 10 kg with an initial speed of 1m s⁻¹ at 45° from a height 1.5 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be 10 m s⁻² , the kinetic energy of the shotput when it just reaches the ground will be

The correct option is 155.0 J.

To find the kinetic energy of the shotput when it just reaches the ground, we can apply the Law of Conservation of Mechanical Energy. Since air resistance is negligible, the total mechanical energy of the shotput remains constant throughout its flight.

The total mechanical energy at any point is the sum of its kinetic energy (due to its motion) and its gravitational potential energy (due to its height above the ground).

Let us write down the given values:
Mass of the shotput, $m=10\text{kg}$
Initial speed, $v_i=1{\text{m s}}^{-1}$
Initial height above the ground, $h=1.5\text{m}$
Acceleration due to gravity, $g=10{\text{m s}}^{-2}$

Step 1: Calculate the initial mechanical energy
The initial kinetic energy ($K_i$) is given by:
$K_i=\frac{1}{2}m{v}_i^2$

Substituting the values:
$K_i=\frac{1}{2}\times 10\times 1^2=5\text{J}$

The initial potential energy ($U_i$) relative to the ground is:
$U_i=mgh$

Substituting the values:
$U_i=10\times 10\times 1.5=150\text{J}$

Therefore, the total initial mechanical energy ($E_i$) is:
$E_i=K_i+U_i=5+150=155\text{J}$

Step 2: Calculate the kinetic energy at the ground
When the shotput just reaches the ground, its height is $0\text{m}$, meaning its final potential energy ($U_f$) is zero:
$U_f=0\text{J}$

According to the conservation of mechanical energy:
$E_f=E_i$

Which means:
$K_f+U_f=E_i$

Since $U_f=0$, the final kinetic energy ($K_f$) is:
$K_f=E_i=155\text{J}$

Note that the angle of projection ($45°$) does not affect the final speed or final kinetic energy when using the work-energy approach because gravity is a conservative force and its work depends only on the vertical displacement.