Physics question suvat

Acceleration is downwards and initial velocity is upwards... so when you're setting up your equations you need to make sure that if the vert component of u is positive the acceleration is negative

I’ve tried this but I’m still not getting the answer:
This was my working:

V^2=8^2+4^2
V=4root5

Then i sued a=v-u/t so -1.62=v-4root5/9 which gives me v as -5.63 which is not an option either?

Thanks!

I don't understand what you did here - but you were on the right track in the first post with v=u+at on the vertical motion, but you've got to set either u or a negative because u and a are vectors pointing in opposite directions.

v=u+at
v=-8+9(1.62)
v=6.58

then by pythagoras

speed = sqrt (6.58^2 + 4^2)

Thanks but I don’t get why you used the vertical component of the initial velocity? I used Pythagoras to find the initial velociy because they have given the horizontal and vertical component and then I plugged it into suvat?

Joinedup is right in that gravity acts downwards so it only affects the vertical component of the motion. So use suvat on the vertical velocity, then combine with the horizontal at the end to get the final speed.

Its worth also noting that you use pythagoras to get the (positive) speed (initial or final), not velocity. Suvat uses velocity as its signed, so irrespective of which direction gravity is acting, you shouldnt be mixing speed and velocity up anyways.

Doesn’t Pythagoras give velocity because they have given horizontal and vertical velocity so the final result would be velocity? Also, what is the purpose of combining the horizontal at the end because we are basically using Pythagoras with the final velocity and initial horizontal component of the velocity? Thanks

Pythagoras is "lengths" so positive values, when you square things up the sign of the legs is irrelevant and sqrt() returns a positive value. So it returns speed. (scalar) Speed here is defined in a 2d space so youd need a direction/angle as well, to determine the (vector) velocity.

Just think about a simple example so a=-1m/s^2 in the vertical direction and consider the two initial velocities
Both have an initial speed of 1m/s. After 1s, their final velocities are (1,-1) and (0,0) respectively, so final speeds of sqrt(2) m/s and 0 m/s. You also know nothing about their direction, even if the speed value is correct.

Ohh ok, I understand. But what’s the point of combining the final velocity with the horizontal velocity, aren’t they two seperate things?

Much of this is gcse physics. Speed (scalar) is the magnitude of the velocity (vector) so its independent of direction. So a speed of 5m/s could correspond to velocities (5,0) or (0,5) or (3,4) or ... it means the blob is covering 5m in 1s, irrespective of the directlion. Its used in the blobs Kinetic Energy as thats 1/2 mv^2 and while v is often referred to as velocity, its really speed and you can think of it as the hypotenuse squared on the right triangle with legs which are horizontal and vertical velocities.

The x and y velocities are independent as the coordinate axes are at right angles, which means you can analyse them seperately, but to work out the KE for instance, you have to combine them to get the blobs speed.