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: 18.9 m
Explanation:
i did the kinematic equation & found the answer.
Answer:
He needs 1.53 seconds to stop the car.
Explanation:
Let the mass of the car is 1500 kg
Speed of the car, v = 20.5 m/s
He will not push the car with a force greater than, 
The impulse delivered to the object is given by the change in momentum as :

So, he needs 1.53 seconds to stop the car. Hence, this is the required solution.
Answer:
–77867 m/s/s.
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 34.5 m/s
Final velocity (v) = –23.9 m/s
Time (t) = 0.00075 s
Acceleration (a) =?
Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:
Acceleration = (final velocity – Initial velocity) /time
a = (v – u) / t
With the above formula, we can obtain acceleration of the ball as follow:
Initial velocity (u) = 34.5 m/s
Final velocity (v) = –23.9 m/s
Time (t) = 0.00075 s
Acceleration (a) =?
a = (v – u) / t
a = (–23.9 – 34.5) / 0.00075
a = –58.4 / 0.00075
a = –77867 m/s/s
Thus, the acceleration of the ball is –77867 m/s/s.
Answer:
C
Explanation:
The pattern is adding .5 to the cm every .1 in weight you just continue the table