# projectile maths question alevel

hello, if anyone could help out with this i’d be very grateful. In this question, take g = 10 ms-2.

A particle P is projected from a point A at the top of a cliff, 52 m vertically above sea level. It moves freely under gravity until it strikes the sea at a point S. The initial velocity components are 24 ms-1 horizontally and 7 ms-1 vertically.

The time of flight of the particle is Answerseconds

The horizontal distance AS is Answerm

The speed of the particle when it hits the sea is
Original post by Superqueen123xxx
hello, if anyone could help out with this i’d be very grateful. In this question, take g = 10 ms-2.

A particle P is projected from a point A at the top of a cliff, 52 m vertically above sea level. It moves freely under gravity until it strikes the sea at a point S. The initial velocity components are 24 ms-1 horizontally and 7 ms-1 vertically.

The time of flight of the particle is Answerseconds

The horizontal distance AS is Answerm

The speed of the particle when it hits the sea is

You need to apply suvat in independent vertical and horizontal directions.
What are you stuck with?
Original post by mqb2766
You need to apply suvat in independent vertical and horizontal directions.
What are you stuck with?

i’ve done that and i know the method but i’m not sure what i’ve done wrong. could you tell me the values of s u v a t to use please
Original post by Superqueen123xxx
i’ve done that and i know the method but i’m not sure what i’ve done wrong. could you tell me the values of s u v a t to use please

It would be good to see / you explain what you tried.
Original post by mqb2766
It would be good to see / you explain what you tried.

so for the first part i worked it out vertically. s as 0. u as 7. a as -10. t is what you find. so i used s is ut+1/2at^2 and got the value of t from that but it’s wrong
Original post by Superqueen123xxx
so for the first part i worked it out vertically. s as 0. u as 7. a as -10. t is what you find. so i used s is ut+1/2at^2 and got the value of t from that but it’s wrong

What is the vertical displacement of the final point (S) relative to the starting point (A)?
Original post by mqb2766
What is the vertical displacement of the final point (S) relative to the starting point (A)?

52. so do you do s as 52 instead of 0??
Original post by Superqueen123xxx
52. so do you do s as 52 instead of 0??

Nearly and Yes.
Is S above or below A? So the sign is ...
Original post by mqb2766
Nearly and Yes.
Is S above or below A? So the sign is ...

below so negative ?
Original post by Superqueen123xxx
below so negative ?

Yes. You've defined up as positive, so down is negative.
So solve the quadratic suvat for time.
Original post by mqb2766
Yes. You've defined up as positive, so down is negative.
So solve the quadratic suvat for time.

is the answer for t 4??
Original post by Superqueen123xxx
is the answer for t 4??

Sub it into the suvat and check it gives s=-52?
Original post by mqb2766
Sub it into the suvat and check it gives s=-52?

it does thanks
hi, i have managed the first part but I can't seem to get the right answer for the speed as it hits the sea. does anyone how to do it? thanks
Original post by Va1kyrie
hi, i have managed the first part but I can't seem to get the right answer for the speed as it hits the sea. does anyone how to do it? thanks

If youve found the time of flight, just sub into the horizontal and vertical velocity suvat equation(s) to calculate the final velocity values, then use pythagoras to combine them.
(edited 2 years ago)
Original post by mqb2766
If youve found the time of flight, just sub into the horizontal and vertical velocity suvat equation(s) to calculate the final velocity values, then use pythagoras to combine them.

Thanks
First to find the time of flight you need to use the vertical distance. S = -52 because the displacement is -52m from the starting point when it hits the sea. It tells you initail velocity in the question, it also tells you to take g =10ms^-1. Using these values in the S = ut + 1/2at^2 equation you should get t = -2.6 or 4. Since t cannot be negative as time cannot be negative then t = 4 seconds.

Next to find the horizontal distance, use the value of t you obtained from the time of flight into the horizontal component. Using equation s = ut you should get 24 x 4 = 96m.

Lastly to find the speed when the particle hits the water you will need to find the final velocity of both the horizontal and vertical distance when time equals 4 and then use pythagorus to get a value of 24^2 + (-33)^2 (root this) --> 40.8ms^-1 to 3.s.f
Original post by Nathaniel911
First to find the time of flight you need to use the vertical distance ...

Its worth having a read over the posting guidelines sticky at the top of the forum. The aim is to give hints to get people to complete as much as possible themselves rather than posting working/solutions. Similarly for not bumping old threads.
Original post by mqb2766
Its worth having a read over the posting guidelines sticky at the top of the forum. The aim is to give hints to get people to complete as much as possible themselves rather than posting working/solutions. Similarly for not bumping old threads.

Oh sorry! How do I make it a spoiler- or should I delete it entirely?
Original post by Nathaniel911
Oh sorry! How do I make it a spoiler- or should I delete it entirely?

Its not a big deal, but Id probably just delete. Note that for instance, there are a few ways to do the question and while the approach you describe is hinted at by the previous question parts, a simple one step solution would just be to conserve energy, so pythagoras at teh start to get u^2 and sub that into
v^2 = u^2 + 2as
as this simply represents
final KE (v^2) = initial KE (u^2) + GPE (2as)
and you dont need to know time and horizontal distance to calculate the final velocity to get the final speed.

Note that as youre innterested in ukmt, you should be able to write down the initial speed u straight away? This kinda hints that this approach was designed into the question.
(edited 1 year ago)