find the minimum velocity for which the range is 39.2m

Asked by shivam upadhyaya | 10th Jun, 2013, 03:55: PM

Expert Answer:

range = v^2 sin2A/g where A is the angle at which projectile is launched and v is the velocity. 
 
v = sqrt(range*g/sin2A)
 
To find the minimum velocity, sin2A = 1
 
v = sqrt(39.2*9.8) = 19.6 m/s

Answered by  | 11th Jun, 2013, 04:20: AM

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