Question

a body is thrown with velocity 20m/s , at what height it's K.E. will be halved​

Answer 1

Answer:

complete the question plz

Answer 2

The height(h) is "10.2 m".

Explanation:

We know that,

Gravitational potential energy  = mgh

Also,

Kinetic energy = $\dfrac{1}{2}$ ×m·$v^{2}$

Given,

Kinetic energy to be halved.

∴ The calculated kinetic energy = $\dfrac{1}{4}$ ×m·$v^{2}$

⇒ mgh = $\dfrac{1}{4}$ ×m·$v^{2}$

 ⇒ gh = $\dfrac{1}{4}$·$v^{2}$

∴ h = $\dfrac{1}{4}$·$v^{2}$ × $\dfrac{1}{g}$

where, g = 9.8 $\dfrac{m}{s^{2} }$

⇒ h = $\dfrac{400}{2(9.8)}$ = 10.2 m

Hence, the height(h) is "10.2 m".