Question

a shot is fired at 30 degree with the vertical from a point on the ground with kinetic energy k if air resistance is ignored the kinetic energy at the top of the trajectory

Answer 1

initially when object is projected at 30 degree with vertical let say its speed is "v"

so the kinetic energy will be

$K = \frac{1}{2}mv^2$

now when object will reach the top position its vertical component of velocity will become zero

It will only have horizontal component of velocity

So at the top position velocity of object will be

$v_f = vsin30 = \frac{v}{2}$

now the kinetic energy will be

$KE' = \frac{1}{2}m(\frac{v}{2})^2$

$KE'= \frac{1}{8}mv^2$

as we compare this to initial kinetic energy

we can say

$KE' = \frac{K}{4}$

so at the top most position its kinetic energy will be one fourth of initial kinetic energy

Answer 2

Answer:

kby8

vsin30

vby2

1/2×m×(v/2)hole square

k/8