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# Physics Projecile Motion Help Please watch

1. A car accelerates and goes over a cliff with a horizontal velocity of 26m/s and falls 80m from the base of the cliff. Find the time taken to hit the ground and the height of the cliff.

I'm hoping someone can solve this because I just can't seem to understand this one. I've been fine with other questions, but this one has been getting me XD
2. (Original post by alfieqe)
A car accelerates and goes over a cliff with a horizontal velocity of 26m/s and falls 80m from the base of the cliff. Find the time taken to hit the ground and the height of the cliff.

I'm hoping someone can solve this because I just can't seem to understand this one. I've been fine with other questions, but this one has been getting me XD
I still cannot understand the question to be honest.

Edit: alfieqe: Does the car reach at the top of the cliff with an initial velocity of 26m/s at the bottom of the cliff?
3. (Original post by alfieqe)
A car accelerates and goes over a cliff with a horizontal velocity of 26m/s and falls 80m from the base of the cliff. Find the time taken to hit the ground and the height of the cliff.

I'm hoping someone can solve this because I just can't seem to understand this one. I've been fine with other questions, but this one has been getting me XD
It's the same as any projectile question. Just sub the relevant values into the relevant suvat formulae.
4. (Original post by morgan8002)
It's the same as any projectile question. Just sub the relevant values into the relevant suvat formulae.
Would you be able to explain the question a bit further as I cannot even understand it

edit: morgan8002: I think I understand it now.
5. If the car has a horizontal velocity of 26m/s and lands 80m from the base of the cliff then to work out time taken is simply distance/speed, 80/26=3.08s

From here you can substitute the values of u (0m/s), a (9.81m/s^2) and t (3.08s) into the equation s = ut + 1/2at^2 to find the height of the cliff
6. (Original post by Mehrdad jafari)
Would you be able to explain the question a bit further as I cannot even understand it?
Maybe a diagram will help. A car drives off a cliff with an initial velocity of 26ms^-1, and falls some distance down and in that time moves 80m across to crash on the ground below the cliff.
7. (Original post by TheHobbit)
If the car has a horizontal velocity of 26m/s and lands 80m from the base of the cliff then to work out time taken is simply distance/speed, 80/26=3.08s

From here you can substitute the values of u (0m/s), a (9.81m/s^2) and t (3.08s) into the equation s = ut + 1/2at^2 to find the height of the cliff
This might be silly but how do you know that the time taken for the car to travel a horizontal distance of 80m is equal to the time taken for the car to fall the vertical distance?
8. (Original post by Mehrdad jafari)
This might be silly but how do you know that the time taken for the car to travel a horizontal distance of 80m is equal to the time taken for the car to fall the vertical distance?
It can be a bit confusing but with projectile questions, horizontal and vertical motion can be treated separately as neither one affects the other. When combined, both components produce the parabolic shape. Since the two components are part of one path of motion, the time taken for both to occur is the same.
Hope this helps
9. most cliffs have water at the bottom...
10. (Original post by TheHobbit)
It can be a bit confusing but with projectile questions, horizontal and vertical motion can be treated separately as neither one affects the other. When combined, both components produce the parabolic shape. Since the two components are part of one path of motion, the time taken for both to occur is the same.
Hope this helps
Yeah, I remember being taught this at AS. Thanks!

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11. (Original post by TheHobbit)
If the car has a horizontal velocity of 26m/s and lands 80m from the base of the cliff then to work out time taken is simply distance/speed, 80/26=3.08s

From here you can substitute the values of u (0m/s), a (9.81m/s^2) and t (3.08s) into the equation s = ut + 1/2at^2 to find the height of the cliff
But surely 'u' doesn't equal 0, because the car accelerates before going off the cliff.
Or maybe im just being dumb
12. (Original post by Alexandra Pett)
But surely 'u' doesn't equal 0, because the car accelerates before going off the cliff.
Or maybe im just being dumb
I think the acceleration here means that the car had to accelerate from rest until reaching the velocity mentioned before going off the cliff because the rate of acceleration is not specified.
13. s=vt s/v=t 80/26= 3.07 seconds i think would seem reasonable
(Original post by alfieqe)
A car accelerates and goes over a cliff with a horizontal velocity of 26m/s and falls 80m from the base of the cliff. Find the time taken to hit the ground and the height of the cliff.

I'm hoping someone can solve this because I just can't seem to understand this one. I've been fine with other questions, but this one has been getting me XD

you get it by doing S=vt for horizontal as the horizontal distance is 80m so, 80 / 26 = 3.076s

Time is the same for both horizontal and vertical.

Use this time and implement it in the vertical component.

Initially the velocity vertically is 0 m/s

S=ut+at^2

0+0.5 x 9.81 x 3.076^2 = 46.41m

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