a stuntman on a motercycle jumps a river which is 5.1m wide. he lands on the edge of the far bank which is 2m lower than the bank from which he takes off.

his minimum horizontal speed is??

it is multi choice paper 2004 question 2

thanks in advance

**0**

# Past paper question please help

Started by Stewartd7, May 23 2009 03:35 PM

3 replies to this topic

### #1

Posted 23 May 2009 - 03:35 PM

### #2

Posted 23 May 2009 - 04:27 PM

Right what you need is the time, the motorcyclist has to cross the gap, and since you know the horizontal distance it is simple to work out the minimum velocity (v=s/t)

to work out the time, you know the vertical displacement (2m) and the acceleration (which is gravity), you can assume his initial vertical velocity is 0 (as he was moving horizontally before going over the edge) so s=ut+(1/2)atē stick in the values and solve...

If you need anymore help just reply

to work out the time, you know the vertical displacement (2m) and the acceleration (which is gravity), you can assume his initial vertical velocity is 0 (as he was moving horizontally before going over the edge) so s=ut+(1/2)atē stick in the values and solve...

If you need anymore help just reply

**=-=-=Marcus=-=-=**### #3

Posted 23 May 2009 - 04:39 PM

QUOTE (Marcus @ May 23 2009, 05:27 PM) <{POST_SNAPBACK}>

Right what you need is the time, the motorcyclist has to cross the gap, and since you know the horizontal distance it is simple to work out the minimum velocity (v=s/t)

to work out the time, you know the vertical displacement (2m) and the acceleration (which is gravity), you can assume his initial vertical velocity is 0 (as he was moving horizontally before going over the edge) so s=ut+(1/2)atē stick in the values and solve...

If you need anymore help just reply

to work out the time, you know the vertical displacement (2m) and the acceleration (which is gravity), you can assume his initial vertical velocity is 0 (as he was moving horizontally before going over the edge) so s=ut+(1/2)atē stick in the values and solve...

If you need anymore help just reply

hi i got t=0.64seconds

so V=5.1 * 0.64=3.264 (i think) but answer is D(8m/s)

### #4

Posted 23 May 2009 - 04:42 PM

QUOTE (Stewartd7 @ May 23 2009, 05:39 PM) <{POST_SNAPBACK}>

QUOTE (Marcus @ May 23 2009, 05:27 PM) <{POST_SNAPBACK}>

Right what you need is the time, the motorcyclist has to cross the gap, and since you know the horizontal distance it is simple to work out the minimum velocity (v=s/t)

to work out the time, you know the vertical displacement (2m) and the acceleration (which is gravity), you can assume his initial vertical velocity is 0 (as he was moving horizontally before going over the edge) so s=ut+(1/2)atē stick in the values and solve...

If you need anymore help just reply

to work out the time, you know the vertical displacement (2m) and the acceleration (which is gravity), you can assume his initial vertical velocity is 0 (as he was moving horizontally before going over the edge) so s=ut+(1/2)atē stick in the values and solve...

If you need anymore help just reply

hi i got t=0.64seconds

so V=5.1 * 0.64=3.264 (i think) but answer is D(8m/s)

forget it im an idiot v+d/t not d*t lol

thanks so much

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