A) What should his speed be at the top of the ramp to make it to the edge of the

Question

A) What should his speed be at the top of the ramp to make it to the edge of the bank? B) if his speed was half the value found in part A where would he land?  A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle (the figure (Figure 1) ). The takeoff ramp was inclined at 53.0 ?, the river was 40.0 m wide, and the far bank was 15.0 m lower than the top of the ramp. The river itself was 100 m below the ramp. You can ignore air resistance.

Explanation / Answer

A) v=17.8 meters/second.

B) L=28.4 meters.

Procedure :

total displacment is 15m....

resolve velocity in x and y
v0x= u cos53.=0.6u
v0y=u sin 53= 0.8u

use eqn x= v0t+0.5at^2 for both vertical and horizontal motion

in y motion
y=v0y*t -0.5*gt^2( let down be -ve so g is -ve)

in xmotion
x=v0x*t (acc in x direction is always 0 cuz velocity in x dirctn is const)

now sub in values
y= -15m( height...-ve cuz its below the starting point)
x=40m( horizontal distance)

now u have 2 eqn in 2 unkowns, u and t
solve simultn...
i got u= 17.8m/s


b) now if his speed is halved....he will land on water...
100m below starting position

so again total displcmt is y=-100 (for vertical motion).....again -ve cuz below strtin point)
they want us to find the length which is in x direction
i.e the hoizontal distance,x

so now that u have u and y...calc t

sub that value for eqn in horizontal motion

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