I have a A level maths question I need ASAP Please

Calculate the time:
*at max height, the velocity is 0 ms^-1

Model gravity downwards

s = /
u = 10 ms^-1
v = 0 ms^-1
a = -9.8 ms^-2
t = ?

v = u + at
0 = 10 - 9.8t
9.8t = 10
t = 10/9.8

Thats for the maximum height yh? thank you

Thats for the maximum height yh? thank you

s = ut + 0.5 at^2

oh ok thank you very much
Can you also help me with the second part "Half a second after the first stone is thrown from the same place with a speed of U ms^-1 at an angle of a(alpha) to the horizontal where tan a=12/5. This stone hits the ground at the same time as the first." please

oh ok thank you very much
Can you also help me with the second part "Half a second after the first stone is thrown from the same place with a speed of U ms^-1 at an angle of a(alpha) to the horizontal where tan a=12/5. This stone hits the ground at the same time as the first." please

oh ok thank you very much
Can you also help me with the second part "Half a second after the first stone is thrown from the same place with a speed of U ms^-1 at an angle of a(alpha) to the horizontal where tan a=12/5. This stone hits the ground at the same time as the first." please

as tan a = 12/5
--> sin a = 12/13 and cos a = 5/13


model horizontal
s = ? [total horizontal distance]
u = U cos a ms^-1
v = /
a = 0
t = time of flight found previous subtract half a second

s = ut

model upwards (till max height which the x-axis is half of total horizontal distance)
s =
u = U sin a ms^-1
v = 0 ms^-1
a = -9.8 ms^-2
t =

I have not gave you the answer for what U is, but the steps above should help you to do so. Feel free to ask if you still do not understand how to get to the answers.

For both horizontal and upwards I’m gonna need to find U yes? Also what happens when I get both total horizontal and upward distance? Is that what I have to use to find U?im going to attemp the question from beginning and try to find U but in the meantime could you send the answer/working for U. Thank you so much btw you have been great help. God bless you

I have not calculated the answer yet.
you should find out the time (t) and max height (s) in terms of U sin a (using model upwards)
and then substitute them together where s = ut + 1/2 at^2

If you cannot find the answer, I will try to calculate

The time would just be 10/9.8 -1 and the max height can be found with the s= ut+1/2at^2. So then I just use the s= ut+1/2at^2 equation to work out U is that right? And when I worked it out I can find the max horizontal distance yeah?

for the max height you can use v^2 = u^2 + 2as

you already previous found out the total horizontal distance by s = ut. you just have to substitute U into the equation

Im sorry I’m very confused now could you send the working out for U, sorry

model upwards (till max height which the x-axis is half of total horizontal distance)
s =
u = U sin a ms^-1
v = 0 ms^-1
a = -9.8 ms^-2
t =


v = u + at
0 = U sin a - 9.8t
9.8t = U sin a
t = (U sin a)/9.8

v^2 = u^2 + 2as
0 = U^2 sin^2(a) - 19.6s
1.96s = U^2 sin^2(a)
s = [U^2 sin^2(a)]/19.6

s = ut + 1/2 at^2
[U^2 sin^2(a)]/19.6 = U sin a [(U sin a)/9.8] + 1/2 (-9.8) [(U sin a)/9.8]^2

I think this is how you do it, so may be able to find U and then sub that into s = ut,
s = (U cos a) (time of flight found previous subtract half a second)

btw, I may not be correct, but tell me as well so I can check my workings