A motorcyclist is trying to jump across a path as shown in Fig. by driving horizontally off a cliff A at a speed of 5ms−1. Ignore air resistance and take g=10ms−2. The speed with which he touches peak B is:
Answers
ANSWER
Speed in horizontal direction remains constant during whole journey because there us no acceleration in this direction .So, v
h =5ms-1
In Vertical direction, loss in gravitation potential energy = gain in KE, i.e.,
mgh= 21 mv
v 2 v v 2 =2gh=2×10×(70−60)=200
Hence The speed with which he touches the cliff B is ;
v= v h 2 +v v 2
= 25+200
= 225
=15ms−1
Step-by-step explanation:
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