Question

A projectile is thrown with a speed 100m/s making an angle 60with horizontly find the tome after which the inclnstion is 45​

Answer 1

We know that in projectile motion, horizontal component of velocity always remains constant in absence of net external force.

Let u be initial velocity and v be the velocity when inclination in 45°.

Horizontal components are ucos60 and vcos45

ucos60=vcos45

100×1/2=v/√2

v=50√2m/s

Now vertical component at this time is vsin45=50√2/√2=50m/s

Initial vertical component=usin60=100×√3/2=50√3m/s

In y direction,

u=50√3m/s

v=50m/s

a=-10m/s^2

t=?

v=u+at

50√3=50-10t

t=50√3-50/10

=3.66s

So time is 3.66s when inclination is 45°

Hope it helps