Average velocity of projectile?

Question

Find the average velocity of projectile betweem the instants it crosses half the maximum height. it is projected at an angle theta with the horizontal at speed u.

The angle with horizontal is "theta" i don't have the symbol here but its well understood i suppose.

Answer 1

Average velocity is distance divided by time.
Between two instants, when projectile crosses half the maximum height, it's displacent is zero
in vertical direction (from +Hmax/2 to +Hmax/2),
and its motion in horizontal direction is uniform.


Answer:
(Uxavg, Uyavg) = (u cosθ, 0)
Uavg = u cosθ

Answer 2

Ok - let ux be the initial speed in x and uy be the initial speed in y then,
ux = u*cos(theta) and uy = u*sin(theta)

The projectile's maximum height is given by conservation of energy, hence

h = uy^2/(2*g) where g = acceleration due to gravity (9.8 m/s^2)

Half that height is h2 = uy^2/(4*g)

Now the projectile's speed at h2 can be found by conservation of energy:

1/2*m*uy2^2 + m*g*uy^2/(4*g) = 1/2*m*uy^2

The masses cancel out and so does a factor of 1/2 so the equation looks like:

uy2^2+uy^2/2 = uy^2

or
uy2^2 = 1/2*uy^2

Finally,
uy2 = uy/sqrt(2)

or uy2 = u*sin(theta)/sqrt(2)

If theta = 45 degress, then sin(45) = sqrt(2)/2 and the result is

uy2 = u/2

Answer 3

No. everyday velocity is entire displacement/time. So in case you throw a ball and after 10s it lands 20m on your suitable, the everyday velocity is 20m/10s = 2m/s to the fabulous. however the on the spot velocity continuously adjustments contained in the direction of the flight. once you throw it, the on the spot velocity while it leaves you hand is various m/s at some perspective to the horizontal.. The magnitude and direction of the on the spot velocity substitute from 2d-to-2d during the flight; through fact the top will enhance the cost decreases and vice versa.

Answer 4

There is only one instant where that occurs, therefore there is no average provided as the solution. The equational answer will produce a definite.

Answer 5

u=hyp
v(y)=u*sin(45) or v(y)=u*sin(135)
v(x)=u*cos(45) or v(x)=u*cos(135)

If i understood the question right !