Help urgent maths

Help can someone explain how to do part 4 b
I got 1.04 but the ans is 2.04 plz explain
I used the maximum distance that I got from part a
Also which suvat do I use

(edited 10 months ago)

Reply 1

Le_Mathemagician

Original post by Alevelhelp.1

Help can someone explain how to do part 4 b
I got 1.04 but the ans is 2.04 plz explain
I used the maximum distance that I got from part a
Also which suvat do I use

You use v = u + at where:
v = 0
u = 10
a = -9.81 (because ball is being thrown up)
t = (find out)

Max height: v = u + at --> 0 = 10 - 9.81t --> rearranging for t will give you t = 10/9.81 = 1.0193... seconds
Given it asks for total time in the air --> t for max height x 2 = 2.0387... seconds = 2.04 seconds (3.sf)

Reply 2

Old man1234

It will decelerate from 10m/s to 0. Om/s being the max height. It will then accelerate from 0m/s. At 9.8/ms2 for the same time again. I can explain further if needed, but I think you have just thought about one direction.

Reply 3

davros

Original post by Alevelhelp.1

Help can someone explain how to do part 4 b
I got 1.04 but the ans is 2.04 plz explain
I used the maximum distance that I got from part a
Also which suvat do I use

you can do it without using part (a) just by using s = ut + (1/2)at^2. You have s= 0 - displacement is 0 because it's returned to where it came from so 0 = 10t + 4.9t^2. t = 0 is one solution to this because that's at the start when it hasn't moved; the other solution is t = 10/4.9 = 2.04 when it returns to its starting position.

Reply 4

Le_Mathemagician

Original post by Old man1234

yes yes I know that already. We both are correct here as there's more than one possible way of answering this question.

Reply 5

Old man1234

Original post by Le_Mathemagician

yes yes I know that already. We both are correct here as there's more than one possible way of answering this question.We posted at the same time. I was not correcting you. No offence intended.