Projectile motion questions help please

[ado21]

Hi, got these as bonus questions but I am having a lot of trouble with answering them, so...

Can anyone help me with these 2?

*Neglecting any air resistance*

1) A body is projected at an angle such that the horizontal range is equal to three times the greatest height. find the angle of projection.

2) A helicopter resting on ground starts to accelerate uniformly upwards. After 30s a flare was fired horizontally from the helicopter. 10s later, the flare reached the ground. (g = 9.81m/s^2).
Calculate:

a)Acceleration of helicopter
b)Height from which flare was fired
c)Height of helicopter the instant flare reaches ground (assuming it continued its constant acceleration after flare was fired)


Thank you so much to whoever answers this.

p.s. they are not supposed to be number answers, variable (i don't think there's any other way)

 

[hage567]

Hi ado21,

You must show what work you've done to solve the questions. We help with homework, we don't do it for you.

 

[baba89]

Hi there, I can definitely help you with these questions. Projectile motion can be a bit tricky, but with some understanding of the basic principles, we can solve these problems together.

For the first question, we need to use the equation for the range of a projectile: R = (v^2*sin(2θ))/g, where R is the range, v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity. In this case, we are given that the horizontal range is equal to three times the greatest height, so we can set up the following equation:

3h = (v^2*sin(2θ))/g

To solve for θ, we need to rearrange the equation to isolate θ. We can do this by multiplying both sides by g and dividing by v^2:

3h * (g/v^2) = sin(2θ)

Now, we can take the inverse sine of both sides to isolate θ:

sin^-1(3h * (g/v^2)) = 2θ

Finally, we can divide both sides by 2 to get our final answer:

θ = sin^-1(3h * (g/v^2))/2

For the second question, we will need to use the equation for the height of a projectile at a given time: h = (v*t) - (1/2*g*t^2), where h is the height, v is the initial velocity, t is the time, and g is the acceleration due to gravity.

a) To find the acceleration of the helicopter, we can use the fact that it is accelerating uniformly upwards. This means that the acceleration is constant, and we can use the following equation: a = (vf - vi)/t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time. We know that after 30 seconds, the flare reached the ground, so we can set vf = 0, and vi = 0 (since the helicopter was initially at rest). This gives us the following equation:

a = (0 - 0)/30 = 0 m/s^2

b) To find the height from which the flare was fired, we can use the equation for the height of a projectile at a given time. We are given that the flare reached the ground after 10 seconds, so we can