I dont get your angle of incline b eing 450?
lets call it theta.
initial vertical speed=100km/hr*1hr/3600sec*1000m/km(*sinTheta=27.7(sinTheta) m/s
initial horizontal speed=27.7(cosTheta)
At the widest gap, he will land at the edge (barely), so jump distance=horizontal speed*timeinair.
timeinair: Hf=hi+vi'cosTheta*time-1/2 g time^2
or time(vi-4.9t)=0 or timeinair= 27.7*sinTheta*/4.9 check that.
widest gap= 27.7*cosTheta*27.7*sinTheta/4.9
if you wish, you can convert sin*cos to another angle using the addition formula, but i see no advantage to doing so.. Note the max width will occur at 45 deg
You must have seen actors in Hindi films jumping over huge gaps on horse backs and motor cycles. In this problem consider a daredevil motor cycle rider trying to cross a gap at a velocity of 100 km h–1. (Fig. 4.13). Let the angle of incline on either side be 450. Calculate the widest gap he can cro
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