First
let us imagine the projectile launched at initial velocity V and at angle
θ relative to the horizontal. (ignore wind resistance)
Vertical component y:
The
initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum
height of h, the vertical velocity
will be 0, therefore the time t taken to attain this maximum height is:
h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g
where
g is acceleration due to gravity
Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike
the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the
horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g]
D = (2V²sinθ cosθ)/g
D = (V²sin2θ)/g
In order for D (horizontal distance) to be
maximum, dD/dθ = 0
That is,
2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2 or θ = π/4
Therefore it is now<span> shown that the maximum horizontal travelled is attained when
the launch angle is π/4 radians, or 45°.</span>
Answer = 6.24x10^18 x ((2 x 3600) + (47 x 60) + 10)
-0 m/s
- average velocity=displacement/time
- the runners displacement is zero so her average velocity must be zero
Answer:
4 m/s
Explanation:
m1 = m2 = m
u1 = 20 m/s, u2 = - 12 m/s
Let the speed of composite body is v after the collision.
Use the conservation of momentum
Momentum before collision = momentum after collision
m1 x u1 + m2 x u2 = (m1 + m2) x v
m x 20 - m x 12 = (m + m) x v
20 - 12 = 2 v
8 = 2 v
v = 4 m/s
Thus, the speed of teh composite body is 4 m/s.
Just because the book is moving doesn't tell you anything about the forces on it, or even whether there ARE any.
Just look at Newton's first law of motion, and this time, let's try and THINK about it too. It says something to the effect that any object continues in constant, uniform MOTION ..... UNLESS acted on by an external force.
