# Horizontal/Vertical projectile question

1. May 10, 2012

### influx

In the mark scheme they have done s/u = t, and then used this t value (along with the vertical component of velocity and g) to calculate the vertical displacement s..

I am confused why they used the same t value to progress with? This t value represents the time taken to travel the 0.4 m (horizontal distance) but surely it would take a different time to travel to the insect?

2. May 10, 2012

### tiny-tim

hi influx!
it follows a parabola (x(t),y(t))

x(t) = ut

so the parabola is (ut,y(t))

you need the point (0.4,y) to lie on that parabola

3. May 10, 2012

### influx

Hi :) !

Ermm, I haven't been taught the notation you are using.. So I don't really understand, sorry lol !

4. May 10, 2012

### azizlwl

It is the same droplet that travels horizontally and vertically.

5. May 11, 2012

### physicist93

First you must resolve horizontally (<--->) to find the time it takes to reach a horizontal distance of 0.4 metres. (You have, s,a and u, so you can find t)

Once you have found this time, t, you must use it to find the vertical height. You do this by resolving vertically. Remember that vertically, u=3.5sin70.

Good luck, hope that helps

6. May 11, 2012

### physicist93

p.s. You can think of that t value as the time it takes for the water droplet to pass/intersect a vertical line 0.4 m away (horizontally) from the fish. It is basically the time it takes for the water droplet to reach the "y" line in your figure.

7. May 11, 2012

### physicist93

p.p.s. I am using online-exam-solutions website for my exam revision. Its pretty good for this sort of thing because they have lots of example questions. Am not sure how relevant it will be for non UK students though...it may be worth checking out

8. May 11, 2012

### influx

Ah! I understand now! Thanks everyone :) Also, I would sign up, but my teacher has given me like 100 questions.. so I am going to have a crack at those! :)