Questions
A particle is projected horizontally as shown in fig. . The values of time of flight and range are
detailed solution
Correct option is A
time of flight for a body that is projected down the inclined plane=T=2usin(α+β)gcosβhere angle of projection wrt horizontal =α=0∘;anglel made by the inclined plane wrt horizontal = β=300;acceleration due to gravity=g=10m/s2;substitute given values in above equationT=(2)(20)sin30∘(10)cos30∘=2.31srange=R=u2gcos2β[sin(2α+β)+sinβ] ; here u= initial velocity of the body projectedα=0∘R=(20)2[sin30∘+sin30∘(10)cos230∘=53.33mTalk to our academic expert!
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