Vertical Motion

[chaput]

Hi there,
Could someone explain the vertical motion concept? I can't seem to get my head around it and there are questions I can't do:(
Also, if you have any great physic videos reccos, please drop them by:)

Thanks

P.S. Questions like these mess with me:( 
A super ball is bounced so that it travels straight up into the air, reaching its highest point after 1.5 s.
a What is the initial speed of the ball just as it leaves the ground? (ans. 14.7 or 15 m s–1 )
b What is the maximum height reached by the ball?  (11.25 or 11 m)

THANK YOU!!

[Sine]

Hi there,
Could someone explain the vertical motion concept? I can't seem to get my head around it and there are questions I can't do:(
Also, if you have any great physic videos reccos, please drop them by:)

Thanks

P.S. Questions like these mess with me:( 
A super ball is bounced so that it travels straight up into the air, reaching its highest point after 1.5 s.
a What is the initial speed of the ball just as it leaves the ground? (ans. 14.7 or 15 m s–1 )
b What is the maximum height reached by the ball?  (11.25 or 11 m)

THANK YOU!!

A super ball is bounced so that it travels straight up into the air, reaching its highest point after 1.5 s.
a What is the initial speed of the ball just as it leaves the ground? (ans. 14.7 or 15 m s–1 )
The highest point is a turning point so v = 0 we also know the time needed so t = 1.5. Gravity is constant so, a= -9.8
v = u + at
0 = u + (-9.8 )(1.5)
0 = u - 14.7
u = 14.7 (m/s)
b What is the maximum height reached by the ball?  (11.25 or 11 m)
so know we know 4 variables
u = 14.7
v = 0
t = 1.5
a = -9.8
so we have 4 equations to choose from and hence can double/triple check your answer
s = 1/2t(v+u)
s = 1/2 * 1.5 (0 + 14.7)
s = 0.75 * 14.7
s = 11.025 (m)

I didn't do VCE physics so may be different more suitable methods

[chaput]

A super ball is bounced so that it travels straight up into the air, reaching its highest point after 1.5 s.
a What is the initial speed of the ball just as it leaves the ground? (ans. 14.7 or 15 m s–1 )
The highest point is a turning point so v = 0 we also know the time needed so t = 1.5. Gravity is constant so, a= -9.8
v = u + at
0 = u + (-9.8 )(1.5)
0 = u - 14.7
u = 14.7 (m/s)
b What is the maximum height reached by the ball?  (11.25 or 11 m)
so know we know 4 variables
u = 14.7
v = 0
t = 1.5
a = -9.8
so we have 4 equations to choose from and hence can double/triple check your answer
s = 1/2t(v+u)
s = 1/2 * 1.5 (0 + 14.7)
s = 0.75 * 14.7
s = 11.025 (m)

I didn't do VCE physics so may be different more suitable methods

Thank you!! This is still really helpful!:)

[KiNSKi01]

To add to this,nearly all the time the vertical journey of an object is comprised of two symmetrical journeys (if that makes sense  )

A helpful equation for free falling objects (applies to an object that has reached its maximum height and begins to descend) is y=1/2 g t^2

Y= displacement g= acceleration t=time.

What Sine said is pretty spot on. For every question right out all the components of suvat/xuvat you know. You will usually have 2 unkown and will need to calculate one of them. After this stage it becomes a process of using the right equation 

Having an actual understanding of the actual physics will definitely be better in the long run; however, strong maths skill alone will still be able to get you the marks!

Hope I helped