Question

a) Deduce an expression for (a) time of flight (b) Horizontal Range (c) Maximum
height reached by projectile in terms of initial velocity and velocity of projection.
b) A projectile is projected from horizontal with velocity v making an angle 45°
with the horizontal direction. Find the distance of the highest point of projectile
from its starting point.​

Answer 1

Answer:

The average velocity of a projectile between the instant it crosses one third the maximum height. It is projected with u making an angle θ with the vertical.

There will be a pair of points for which vertical velocities at the same height are in opposite direction and therefore their average sum =0

It is the horizontal velocity which is uniform and hence v

av

=u

x

=ucosθ

For a general point:

Displacement in Ye direction:

y=usinθ×t−

2

gt

2

Displacement in X-direction:

x=ucosθ×t

Now in order to calculate average velocity:

Average Velocity =

Total time

net displacement

Explanation:

please follow me

Answer 2

Answer:

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