first we make a U turn and travel towards home in t = 20 s
so the distance of home from initial position is
Now after picking up the book we travel back with speed 20 m/s
so again after t = 20 s the displacement is given as
so the net displacement is given as
so it will be displaced by total displacement 200 m
Answer:
The last part on the right side of the diagram
Explanation:
Im on plato and just got it right :)
Answer:
(a) The range of the projectile is 31,813.18 m
(b) The maximum height of the projectile is 4,591.84 m
(c) The speed with which the projectile hits the ground is 670.82 m/s.
Explanation:
Given;
initial speed of the projectile, u = 600 m/s
angle of projection, θ = 30⁰
acceleration due to gravity, g = 9.8 m/s²
(a) The range of the projectile in meters;
(b) The maximum height of the projectile in meters;
(c) The speed with which the projectile hits the ground is;
Answer: Both cannonballs will hit the ground at the same time.
Explanation:
Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.
then the acceleration equation is only on the vertical axis, and can be written as:
a(t) = -(9.8 m/s^2)
Now, to get the vertical velocity equation, we need to integrate over time.
v(t) = -(9.8 m/s^2)*t + v0
Where v0 is the initial velocity of the object in the vertical axis.
if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s
and:
v(t) = -(9.8 m/s^2)*t
Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)
And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.
You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)