The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22
1 answer:
1) The projectile's motion follows
,
In order to determine the velocity, we must compute the derivative of h(t): Next, we will compute the speed at t=2 s and t=4 s: The negative value of the second speed suggests that the projectile has already attained its highest point and is now descending. 2) The maximum height of the projectile occurs when its speed equals zero: Thus, we have And solving yields
3) To determine the maximum height, we substitute the time at which the projectile reaches this peak into h(t), specifically t=2.30 s: 4) The time at which the projectile lands is when the height reaches zero; h(t)=0, which leads to This results in a second-degree equation, producing two answers: the negative root can be disregarded as it lacks physical significance; the second root is
, which indicates the landing time of the projectile. 5) The moment the projectile impacts the ground corresponds to the velocity at time t=4.68 s:
, carrying a negative sign to denote a downward direction.
,
In order to determine the velocity, we must compute the derivative of h(t): Next, we will compute the speed at t=2 s and t=4 s: The negative value of the second speed suggests that the projectile has already attained its highest point and is now descending. 2) The maximum height of the projectile occurs when its speed equals zero: Thus, we have And solving yields
3) To determine the maximum height, we substitute the time at which the projectile reaches this peak into h(t), specifically t=2.30 s: 4) The time at which the projectile lands is when the height reaches zero; h(t)=0, which leads to This results in a second-degree equation, producing two answers: the negative root can be disregarded as it lacks physical significance; the second root is
, which indicates the landing time of the projectile. 5) The moment the projectile impacts the ground corresponds to the velocity at time t=4.68 s:
, carrying a negative sign to denote a downward direction.
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Answer:
The work done, W = 19.6 J
Explanation:
It’s provided that
The mass of the block, m = 5 kg
The velocity of the block, v = 10 m/s
The coefficient of kinetic friction between the block and rough surface is 0.2
Distance traveled by the block, d = 2 m
As the block traverses the rough section, it loses energy equal to the work done by the kinetic energy.
W = 19.6 J
Thus, the change in kinetic energy of the block moving through the rough section is 19.6 J. Consequently, this is the required answer.
Answer:
a. Angle= 28.82°
b. Approved. Although he might feel cold, he should be able to cross.
Explanation:
Velocity Vector
Velocity is a measure of how quickly something is moving in a specific direction. It is represented as a vector that has both magnitude and direction. If an object can only move in one direction, then speed can serve as the scalar equivalent of that velocity (only focusing on magnitude).
a.
The explorer aims to swim across a river to reach his campsite, as depicted in the image below. The river's velocity is vr and the explorer's swimming speed in still water is ve. If he were to swim straight towards the campsite, he would end up downstream due to the river's current. Therefore, he must swim at an angle that allows him to overcome the current while still moving towards his goal. This angle relative to the shore is what we need to determine. The explorer's speed can be broken down into its horizontal (vx) and vertical (vy) components. In order to counteract the river's flow:
We can calculate the vertical component of the explorer's swimming speed as
Thus
Finding the value of
Then the angle is given by
b.
The component of the explorer's velocity that goes horizontally is
This represents the actual velocity directed towards the campsite
Considering that
To find t
Calculating the duration for the explorer to cross the river
As this time is under the hypothermia threshold (300 seconds), the conclusion is
Approved. Although he will feel cold, he should manage to cross successfully.
