Give answers of all these questions.
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Answers
1) Yes velocity can be in reverse direction which is termed as negative velocity for example, the bouncing back of a cricket ball.
2) the ball which is thrown upward will hit the ground with higher speed due to more height and gravitation the kinetic energy increases, that's why.
3) Yes, the statement is true.
By definition, acceleration means change in velocity. So if the velocity is constant, this means both speed and direction are constant. And there has to be at least one of these things changing for the velocity to change, the acceleration is zero if none of these things change and the velocity is said to be constant.
4) Impulse is the change of momentum of an object when the object is acted upon by a force for an interval of time. So, with impulse, you can calculate the change in momentum, or you can use impulse to calculate the average impact force of a collision. The formula for impulse is: Impulse = Force * time = force * Delta t.
Momentum is mass in motion, and any moving object can have momentum. An object's change in momentum is equal to its impulse. Impulse is a quantity of force times the time interval. Impulse is not equal to momentum itself; rather, it's the increase or decrease of an object's momentum.
5) Head-on collisions happen when two cars driving in opposite directions crash into each other. Although they are rare, they can be devastating for both parties, and they may even result in one or more wrongful deaths. They may also happen because of drunk driving, distracted driving, or brake failure.
6) At any time t, a projectile's horizontal and vertical displacement are:
x = VtCos θ where V is the initial velocity, θ is the launch angle
y = VtSinθ – ½gt^2
The velocities are the time derivatives of displacement:
Vx = VCosθ (note that Vx does not depend on t, so Vx is constant)
Vy = VSinθ – gt
At maximum height, Vy = 0 = VSinθ – gt
So at maximum height, t = (VSinθ)/g
The range R of a projectile launched at an angle θ with a velocity V is:
R = V^2 Sin2θ / g
The maximum height H is
H = V^2 Sin^2(θ) / 2g
So the ratio of range R to height H is:
R/H = [V^2 Sin2θ / g] / [V^2 Sin^2(θ) / 2g] = 2Sin2θ / Sin^2(θ)
= 2(2SinθCosθ)/Sin^2θ = 4Cosθ/Sinθ = 4Cotθ
Therefore, the range of a projectile can certainly exceed 4 times the height, since R/H = 4Cot(θ) has a value of +∞ to -∞.
7) If the angle of projection is 75.96 degrees the maximum height is equal to the horizontal range.
8) When released, the bombs are moving horizontally the same speed as the aircraft that releases them, however, wind resistance will tend to slow down their horizontal movement, however, they will still be moving the same direction horizontally as the bomber was going when the bombs were released. How far they travel horizontally from release to impact depends on the speed of the aircraft, the orientation of the aircraft (level, in a dive or in a climb) and the height above the ground the bombs were dropped. Other factors include the wind direction and strength and the type of bomb used. Some are designed to be high drag and some are designed to be low drag.
Answer:
hey mate here is your answer
Explanation:
Any object that is in flight after being thrown or projected is known as projectile motion