Question

A projectile is thrown with the velocity v at an angle with hozrizontal when the projectile

Answer 1

A projectile is thrown with velocity v at an angle  Theta with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is? A projectile is thrown with velocity v at an angle Θ with horizontal.

Answer 2

If the point of projection is considered the origin, then the co-ordinates of the highest point of the trajectory of the projectile would be (R/2,H), where R is the Range of the projectile and H is the maximum height of the projectile.

From corresponding formulae, R/2=[(v^2)sin(2*theta)]/2g and H=[(v^2)(sin(theta))^2]/2g. The time taken to move from the point of projection and the top-most point is half of the time of flight. T/2=[vsin(theta)]/g.

Thus, the average velocity vector of the projectile between the point of projection and the highest point is given by two-point form as:

x component= x component of displacement/time

=(R/2)/(T/2)=R/T= 2vcos(theta)

y component= y component of displacement/time

=H/(T/2)=2H/T= vsin(theta)

Thus, final average velocity vector is given by v= 2vcos(theta)i+ vsin(theta)j

MARK BRAINLIEST..