Question

a projectile whose masss is 10g is fired directly upward from ground level with an initial velocity of 1000m/s neglecting the effects of air resistance what will be the speed of the projectile when it impacts the ground

Answer 1

To find the speed of the projectile when it impacts the ground, we need to consider the influence of gravity. The projectile will continue to move upward until its velocity becomes zero, and then it will start falling back down.

Step-by-step explanation:

In this problem, a projectile with a mass of 10g is fired directly upward from ground level with an initial velocity of 1000m/s, neglecting air resistance. To find the speed of the projectile when it impacts the ground, we need to consider the influence of gravity. The projectile will continue to move upward until its velocity becomes zero, and then it will start falling back down. The time it takes for the projectile to reach its maximum height can be determined using the equation:

t = (2 * initial vertical velocity) / acceleration due to gravity

After finding the time, we can calculate the maximum height the projectile reaches using the equation:

maximum height = initial vertical velocity * t - (1/2) * acceleration due to gravity * t^2

Finally, we can calculate the speed of the projectile when it impacts the ground using the equation:

speed = initial vertical velocity + acceleration due to gravity * t

Plugging in the given values, we can find the speed at which the projectile impacts the ground.

Answer 2

The speed of the projectile when it impacts the ground is 1000 m/s

Step-by-step explanation:

Vertical Launch

When an object is launched vertically and upwards it starts to move at an initial speed vo, then the acceleration of gravity makes that speed to reduce until it reaches 0. The object has reached its maximum height. Then, it starts to move downwards in free fall, with initial speed zero and gradually increasing it until it reaches the ground level. We will demonstrate that the speed it has when impacts the ground is the same (and opposite) as the initial speed vo.

The speed when the object is moving upwards is given by

`v_f=v_o-g.t`

The time it takes to reach the maximum height is when vf=0, i.e.

`0=v_o-g.t`

solving for t

`\displaystyle t=(v_o)/(g)`

The maximum height reached is

`\displaystyle y=(gt^2)/(2)=(v_o^2)/(2g)`

Then, the object starts to fall. The object's height is given by

`\displaystyle y=(v_o^2)/(2g)-(gt'^2)/(2)`

where t' is the time the object has traveled downwards. The height will be 0 again when

`\displaystyle (v_o^2)/(2g)=(gt'^2)/(2)`

Solving for t'

`\displaystyle t'=(v_o)/(g)`

We can see the time it takes to reach the maximum height is the same it takes to return to ground level. Of course, the speed when it happens is

`v_f=g.t'=v_o`

Thus, the speed of the projectile when it impacts the ground is 1000 m/s