Question

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

Answer 1

Answer:

155 J

Explanation: Change in Mechanical Energy is zero.

                      ΔM.E. = 0

ΔK.E+ΔP.E=0

ΔK.E = -ΔP.E

See attachment for solving

Answer 2

Given:

mass of the shot put , m= 10 Kg

initial speed , u = 1 m/s

throwing angle = 45°

height , h = 1.5 m

g = 10 m/$s^{2}$

To find :

Kinetic energy of the shot put when it just reaches the ground .

Solution :

By conservation of energy ,

total initial energy = total final energy

→ ( initial kinetic energy + initial potential energy ) = ( final kinetic energy + final potential energy )

→ ( 0.5 *m*u*u ) + m*g*h = final kinetic energy + 0

→ ( 0.5 * 10 * 1 * 1) + ( 10 * 10 * 1.5 ) = final kinetic energy

→ 5 + 150 = final kinetic energy

→ final kinetic energy = 155 J

Kinetic energy of the shot put when it just reaches the ground is 155 J