Suvat equation question - projectiles

The cliff divers of Acapulco, Mexic, take off from a rocky cliff fact 26.5 m above the surface of the water. In the course of their flight, of which all but the last 1.5 of their horizontal motion is above rock, they travel 8.0 m forward.

For how long are they in the air (ignoring air resistance)
For how long are they over the water during their flight.

Firstly i'm not too sure the difference in the 2 questions, but more importantly I can only come up with 2 unknowns for either the horizontal/vertical.

Horizontal I know u and s
Vertical a and s

any help?

Reply 1

F1 fanatic

16

Horizontally you have constant velocity. So the equation reduces to s=vt. That gives you the time (first bit), which also gives you the 3rd variable for the vertical bit. however, you say all but the last 1.5... 1.5 what? no units so is it time or distance?

btw, you also know the vertical a, it's just the acceleration due to gravity... or did you get vertical and horizontal confused?

Reply 2

cvat

OP

F1 fanatic

Horizontally you have constant velocity. So the equation reduces to s=vt. That gives you the time (first bit), which also gives you the 3rd variable for the vertical bit. however, you say all but the last 1.5... 1.5 what? no units so is it time or distance?

btw, you also know the vertical a, it's just the acceleration due to gravity... or did you get vertical and horizontal confused?

1.5 is in metres, and for horizontal how would you work out v when no speed is given and yeah I got horizontal and vertical the wrong way round gonna edit

Reply 3

teachercol

9

Work out the vertical time of flight to cover 26.5m. using suvat

In this time, diver moves 8.0m forward, so can now work out horiz velocity

Reply 4

cvat

OP

teachercol

Work out the vertical time of flight to cover 26.5m. using suvat

In this time, diver moves 8.0m forward, so can now work out horiz velocity

i don't really understnad what you mean.

Firstly from what i've gathered from the question
'In the course of their flight, of which all but the last 1.5m of their horizontal motion is above rock, they travel 8.0 m forward.'

Therefore the first 6.5m are above the height of the cliff therefore I don't have s (distance) as I do not know how high he actually jumped above the height of the cliff. If you get what I mean?

Reply 5

F1 fanatic

16

cvat

i don't really understnad what you mean.

Firstly from what i've gathered from the question
'In the course of their flight, of which all but the last 1.5m of their horizontal motion is above rock, they travel 8.0 m forward.'

Therefore the first 6.5m are above the height of the cliff therefore I don't have s (distance) as I do not know how high he actually jumped above the height of the cliff. If you get what I mean?

you assume he doesnt jump up at all, that he just runs off it. My reading of it is that the "rocks" are below, and not part of the cliff themselves. It's a rocky shore below if you like. Does that help?

Reply 6

teachercol

9

The cliff divers of Acapulco, Mexic, take off from a rocky cliff fact 26.5 m above the surface of the water

Work out the vertical time of flight to cover 26.5m. using suvat

s = 0.5 gt^2 assuming no initial vertical velocity ie diver takes off horizontally.

Reply 7

cvat

OP

F1 fanatic

you assume he doesnt jump up at all, that he just runs off it. My reading of it is that the "rocks" are below, and not part of the cliff themselves. It's a rocky shore below if you like. Does that help?

yep that's exactly what I misunderstood, we went through it today at school, should of thought of it but there was a picture on the question before about a long jumper jumping up first, which clouded my judgement, thanks for all the help though