InterviewSolution
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InterviewSolution
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A cricket fielder can throw the cricket ball with a speed ` v_0`. If he throws the ball while running with speed (u) at angle ` theta` to the horizontal, find (b) what will be time of flight ? (c ) what is the distance (horizontal range) form the point of projection at which the ball will land ? (d) find ` theta` at which he should throw the ball that would maxmise the horizontal range range as found in (c ). (e) how does ` theta` for maximum range change if ` u gt v_0, u=v_0, ult v_0` ? (f) how does ` theta` in (e) compare with that for ` u=0 (i.e., 45^@) ?A. `theta_(max)=cos^(-1)[(-u+sqrt(u^(2)+8v_(0)^(2)))/(4v_(0))]`B. `theta_(max)=cos^(-1)[(u+sqrt(u^(2)+8v_(0)^(2)))/(4v_(0))]`C. `theta_(max)=cos^(-1)[(-u+sqrt(u^(2)-8v_(0)^(2)))/(4v_(0))]`D. `theta_(max)=cos^(-1)[(-u+sqrt(u^(2)-4v_(0)^(2)))/(4v_(0))]` |
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Answer» Correct Answer - A For horiontal range to be maximum,`(dR)/(dtheta)=0` `rArr theta_(max)=cos^(-1)[(-u+sqrt(u^(2)+8v_(0)^(2)))/(4v_(0))]` |
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