Urgent help needed Physics 9702/P1/M/J/2004 Q9

[ArsalanT]

A motorcycle stunt rider moving horizontally takes off from a point 1.25 m above the ground, landing 10 m away as shown.

What was the speed at take off?

A.5 m/s B. 10 m/s C. 15 m/s D. 20 m/s

SUMEONE PLZ REPLY URGENTLY WITH PROPER EXPLANATION I'LL BE REALLY GRATEFUL... THNX

 

[beacon_of_light]

Thats just like a parabola path but a little modified ... we have to find the horizontal speed at take off... use second equation of motion ...

s = 1/2 gt^2
1.25 = 1/2 x 9.8x t^2
find time of flight
t = 0.51s
use speed = distance /time
= 10/0.51 = 20m/s

hope you get that!

 

[ArsalanT]

ohh!!....i dint get that.....nywayz thanks a lot!!.

 

[ArsalanT]

i have got another question if u dont mind.............plz reply for this also!!....

O/N/2004 P1 Q 18,,,,,,
A cyclist is capable of generating an average power of 3kW during a 4km speed trial. His aerodynamic suit and position on the cycle reduce resistive forces to 180N.
What is the approximate time achieved in the speed trial?

A. 140s B. 240s C. 1300s D. 2200s

 

[larina]

thats oct/nov 04 question no. 17...not 18

well here...power is the rate of energy(work done)
so power= energy(work done)/time
so power=(force x distance)/time
the power should be in Watts, force in Newtons, Distance in meters and time in seconds.

so here...they're asking us to find the time....
so time= (force x distance)/ power
force-----> 180 N
distance-->4km = 4000 m
power---->3kW = 3000W

so simply (180 x 4000)/ 3000
= 240 s
so it's (B)

 

[creative22]

thats oct/nov 04 question no. 17...not 18

well here...power is the rate of energy(work done)
so power= energy(work done)/time
so power=(force x distance)/time
the power should be in Watts, force in Newtons, Distance in meters and time in seconds.

so here...they're asking us to find the time....
so time= (force x distance)/ power
force-----> 180 N
distance-->4km = 4000 m
power---->3kW = 3000W

so simply (180 x 4000)/ 3000
= 240 s
so it's (B)

hahah it seems easy but the answer is( A ) in the marking scheme

what i think is that as the resistance (air resistance ) is decresing the force of motion should incraese and so is the power and hense time decrease than 240 s and that is only in ( A ) part

 

[larina]

no..the question is 17 not 18.
and the answer is B to it

 

[creative22]

no..the question is 17 not 18.
and the answer is B to it

ohh now i got you
it was easy question by the way
ARSLAN confused us

 

[beacon_of_light]

ohh!!....i dint get that.....nywayz thanks a lot!!.

You should go through parabolic motion once thoroughly...

 

[ArsalanT]

@larina and creative........ in the o/n/04 question it says the resistive force = 180 N....... so the effective force should be (F-180).....now we have 2 variables (F and t) ....how can we find out time!!..i used E=Pt......(F-180) x 4000 = 3000t............i am unable to make any other equation in terms of F and t.......

 

[grumpy]

its simple it says "average power" not just power.so it means that there is an average velocity.if velocity is constant then it means resitive force=driving force.
then you can simply calculate.

 

[Bimal]

The time taken by the stunt-man to reach the given distance is equal to the time taken by him to fall horizontally to the ground.now by using
s=ut+1/2at^2
putting u=o m/s and a=10 m/s^2
we get t=0.5 sec
now,velocity(horizontal)=(horizontal distance)/time
=10/0.5
=20m/s which is the initial velocity since horizontal velocity is always constant.

 

[Rajendra bogati]

at first calculate time taken: t^2=2H/G ; H=1.25m and then use the formula X=UT where X is the horizontal distance i.e. 10m and put the value of T then u will got the value of U(the speed of take-off)... got it?

 

[Amrah Salman55]

Thats just like a parabola path but a little modified ... we have to find the horizontal speed at take off... use second equation of motion ...

s = 1/2 gt^2
1.25 = 1/2 x 9.8x t^2
find time of flight
t = 0.51s
use speed = distance /time
= 10/0.51 = 20m/s

hope you get that![/QUOTE
so helpful thanks alot