Question

A projectile is thrown in the upward direction making an angle of 60∘ with the horizontal direction with a velocity of 150ms−1. Then the time after which its inclination with the horizontal is 45∘ is...........please reply

Answer 1

At the two points of the trajectory during projectile motion, the horizontal component of the velocity is same. Then

150×12=v×12–√150×12=v×12 or v=1502–√ms−1v=1502ms-1

Initially : uy=usin60∘=1503–√2ms−1uy=usin60∘=15032ms-1

Finally : vy=vsin45∘=1502–√×12–√=1502ms−1vy=vsin45∘=1502×12=1502ms-1

But vy=uy+aytvy=uy+ayt or 1502=1503–√2−10t1502=15032-10t

10t=1502(

$\sqrt{3 - 1}$