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A target is fixed on top of a pole 13 m high. A person standing at a distance 50 m from the pole is capable of projecting a stone with velocity 10*rootg(underroot 9.8 m/s^2). If his aim is to strike the target in least time what should be the angle of elevation with time

A target is fixed on top of a pole 13 m high. A person standing at a distance 50 m from the pole is capable of projecting a stone with velocity 10*rootg(underroot 9.8 m/s^2). If his aim is to strike the target in least time what should be the angle of elevation with time

Grade:11

2 Answers

Arun Kumar IIT Delhi
askIITians Faculty 256 Points
6 years ago
Hello Student,
y=13=vsintheta*t-1/2gt^2
x=50=vcostheta*t
we need to find dtheta/dt
from 1stequation
0=vcostheta*t*dtheta/dt+vsintheta-gt
from second eq.
0=-vsintheta*t*dtheta/dt+vcostheta
we need to do dtheta/dt=0
=>vcostheta=0 but it isn’t possible
and vsintheta-gt=0
=>sintheta=gt/v
Thanks & Regards
Arun Kumar
Btech, IIT Delhi
Askiitians Faculty
muktesh singh
16 Points
6 years ago
Sir how to calculate time, please give detailed solution to this question.
 

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