If a thunder clap comes 1.5 s after the lightning strike, how far away is thestrike?
1 answer:
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Consider the projectile launched at initial velocity V at angle θ relative to the horizontal.
Neglect wind or aerodynamic resistance.
The initial vertical velocity is Vsinθ.
When the projectile reaches its maximum height of h, its vertical velocity will be zero.
If the time taken to attain maximum height is t, then
0 = Vsinθ - gt
t = (Vsinθ)/g, where g = acceleration due to gravity.
The horizontal component of launch velocity is Vcosθ. This velocity remains constant because aerodynamic resistance is ignored.
The time to travel the horizontal distance D is twice the value of t.
Therefore
D = Vcosθ*[(2Vsinθ)/g]
= (2V²sinθ cosθ)/g
= (V²sin2θ)/g
In order for D (horizontal distance) to be maximum,
That is,
Because
, therefore cos(2θ) = 0.
This is true when 2θ = π/2 => θ = π/4.
It has been shown that the maximum horizontal traveled can be attained when the launch angle is π/4 radians, or 45°.
Neglect wind or aerodynamic resistance.
The initial vertical velocity is Vsinθ.
When the projectile reaches its maximum height of h, its vertical velocity will be zero.
If the time taken to attain maximum height is t, then
0 = Vsinθ - gt
t = (Vsinθ)/g, where g = acceleration due to gravity.
The horizontal component of launch velocity is Vcosθ. This velocity remains constant because aerodynamic resistance is ignored.
The time to travel the horizontal distance D is twice the value of t.
Therefore
D = Vcosθ*[(2Vsinθ)/g]
= (2V²sinθ cosθ)/g
= (V²sin2θ)/g
In order for D (horizontal distance) to be maximum,
That is,
Because
This is true when 2θ = π/2 => θ = π/4.
It has been shown that the maximum horizontal traveled can be attained when the launch angle is π/4 radians, or 45°.
Answer:
Magnitude 900m/s, direction 12.8° respect to the velocity of the first asteroid.
Explanation:
This is a perfectly inelastic collision, because the two asteroids stick together at the end. That means that the kinetic energy doesn't conserves, but the linear momentum does. But, since the velocities of the asteroids have different directions, we have to break down them in components. For convenience, we will take the direction of the first asteroid as x-axis, and its perpendicular direction (in the plane of the two velocity vectors) as y-axis. So, we have that:
And, since , we get:
Solving for v_fx and v_fy, and calculating their values, we get:
Now, the final speed can be calculated using the Pythagorean Theorem:
And the direction can be obtained using trigonometry:
That means that the final velocity of the two asteroids has a magnitude of 900m/s and a direction of 12.8° with respect to the velocity of the first asteroid.