Help with Projectile Motion (1 Viewer)

[locked.on]

I'm having some difficulties interpreting the question.

In my attempts at the question I have drawn out several diagrams and resolved the vertical and horizontal components for each . But I can't seem to acquire the same answers provided by the solution no matter which diagram.

Any help on achieving the final solution is much appreciated!

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A cyclist wants to perform a stunt in which he rides up a ramp, launching himself into the air, then flies through a hoop and lands on another ramp. The angle of each ramp is [30.0 degrees] and the cyclist is able to reach the launch height of 1.50m with a launching speed of 30.0km/h.

1. At what maximum height above the ground could the lower edge of the hoop be placed?
2. How far away should the landing ramp be placed?

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Thanks!
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[alez]

i get 15 for initial vertical velocity
and 25.98076211 for horiz

1. so v^2=u^2+2as
0=15^2+2x9.8xs
s=11.47959184
s+1.5=12.97959184

2. then to find t
v=u+at
0=15-9.8t
t=1.530612245
total time=tx2=3.06122449

s=ut+1/2at^2
s=3.06x30cos30
s=79.53294525

is that right?

 

[locked.on]

Not quite - remember you have to change km/h into m/s as well.

 

[alez]

oh shit yea. so v=8 1/3
vertical u=4 1/6
horiz u=7.216878365
is that right before i do it all again?

 

[alez]

1.s=2.386
2.t=0.85
s=6.14

 

[locked.on]

Those components are the same as mine, except I can't get out the solutions from the book.

Thanks for your efforts.

 

[alez]

well the book might be wrong
like the surfing books are bullshit
reliability is on our side

 

[locked.on]

Alright - I'll put up my solutions and the books.

EDIT;
Actually I take it our solutions are correct.

Thanks for your time and effort - good luck for your HSC!

 

[henry08]

This is from the Jacaranda textbook isn't it?
Here's my worked solution:

 

[locked.on]

It is from Jacaranda, well picked up.

Thanks for your working - really helped with setting out.