Question

A projectile is projected at an angle 60 degree from the horizontal calculate its kinetic energy and the maximum height if its initial kinetic energy is E

Answer 1

Answer:

maximum height

$h = \frac{ {u}^{2} { \sin(60) }^{2} }{2 \times g}$

or,

$h = \frac{ {u}^{2} 3}{g8}$

initial kinetic energy=E=

$\frac{1}{2} \times m \times {u}^{2}$

when u is the initial velocity and m is the mass of particle.

then the kinetic energy at the highest point,

$= \frac{1}{2} \times m \times {u }^{2} { { \cos(60) }^{2} }$

=E/4

Answer 2

Answer:

velocity at maximum height;Vx=u cosalpha

initial k.e=1/2mu2

k.e at maximum height E1=1/2mVx2=1/2mu2cos2alpha

=1/2 cos(60)^2=E/4