Projectile motion question

I have the above question which I need some help with.

I have tried two different ways to get a. Both methods used the formula
V² = U² + 2as

Making U = 0 and a = -9.81

As for S I tried using trig to find the hypotenuse of the triangle formed which then gave me a final answer of 39.9m/s. I also just tried plugging the height of 55m to the above formula but that gave me 32 m/s, the answer I need is 37 m/s.

I know how to do B once I have the V value, I can then find the flight time for both halves of the flight curve and find the distance from that, im just stuck on how to get started.

Reply 1

Stonebridge

Calculate the time it would take and the initial velocity required if the object were projected vertically upwards, to reach a height of 55m. (=90-35)
Uses basic SUVAT formulas.
This then gives you the time to reach the top of the wall because this would be at the highest point of the trajectory. It also gives you the required vertical component of the initial velocity.

Next step.

Spoiler

Reply 2

username931278

so using the above formula, would I get it like this

V = 2 x -9.81 x 55 = 32m/s ?

Reply 3

Stonebridge

Original post by Diggedy

so using the above formula, would I get it like this

V = 2 x -9.81 x 55 = 32m/s ?

Should have been v² but yes,
That gives the vertical component of the initial velocity.
The spoiler in my previous post tells you how to get the horizontal component, and thus the actual velocity and the angle.

Reply 4

username931278

ok, ive then use V = u + at to find the time which is 3.34second.

However when plugging this into s = 0.5 (u + v) x t to find the maximum height I only get 54.7m from origin, which says to me it wouldnt clear the wall so im assuming theres something wrong in my figures.

Also, how do I find the angle from this?

Reply 5

Stonebridge

Original post by Diggedy

Why are you trying to find maximum height? It's given in the question as 55m and you used it to find the initial vertical component. (By the way, I get it t be 32.8m/s so check your arithmetic. It may depend on the value of g you use.)
You've plugged your answer back in to find what you were given initially. The reason it's not the same is rounding errors.

The time value is correct.

The angle is found from the fact that its tangent is equal to the ratio of the vertical and horizontal components.
Have you calculated the horizontal component from the distance to the wall and the time taken to get there?

Reply 6

username931278

sorry, I was getting myself lost.

What ive now done is worked out the hypotenuse of the triangle to be 108.16. Then ive divided that length by 32.84 then multiplied it 60 to get my x velocity which is 18.27 m/s.

Ive then used the two velocities to get the angle of 56.3 degrees.

Then ive used the overall height to work out the second half of the flight time to be 5.48. combined the two times and multiplied it by the x velocity to give me an overall distance of 160.8 meters.

Is this correct?

The answers I have on the sheet are Vy = 37.42 m/s, 61.4 degrees, 136.7 metres.

Reply 7

Stonebridge

So we have the vertical component of the initial velocity is 32.8m/s (from consideration of the height of 55m needed to clear the wall)
We also have that the time taken to get to this height (v=u+gt) is 3.34s
This time means you can find the (constant) horizontal component from
horizontal distance to wall = horizontal component x time (basic s=vt)
This gives you 60/3.34 for the velocity. This is 18m/s approx

The actual projected velocity is V and given by

V^2 = V^2_{horizontal} + V^2_{vertical}

and comes to 37.4m/s

The angle is

tan^-1( \frac{vertical component}{horizontal component})

and comes to 61 degs approx.

The overall time will be the initial 3.34s plus
the time it takes the object to fall from a height of 90m to the ground. (basic SUVAT on the vertical motion.)
The overall horizontal distance will be this time x horizontal component of velocity.

(edited 11 years ago)

Reply 8

username931278

so is it a fluke that my method of finding the x component worked, or is that also a valid method?

Reply 9

Stonebridge

I don't understand your method. What triangle and what hypotenuse?
The only triangle and hypotenuse you need is to find the actual projected velocity from the two components.
As the angle is incorrect by your method I suspect it's not valid.
I suggest doing the question using my method. It's systematic and gives the right answers.
The technique is always to consider the vertical and horizontal motion separately.
Times are found from the vertical motion using SUVAT equations and the other info is found from the horizontal motion using those times.

Reply 10

username931278

ah, sorry I guess I confuse myself when explaining things

What I did was use sqrt 60² + 90² = 108.16

The I did (32.8/108.16) x 60 = 18.19 (x component) so it was creating a triangle to give me a ratio to work the components from.

had I then used these two values I would have found the actual value of V, then I could have worked out the angle.

This method gave me the same value of V = 37.4, which would have also allowed me to get the correct angle.

However, thanks for your help!