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OpenStudy (anonymous):

the range of projectile when launched from ground at an angle of 15degree with the horizontal is 1.5km the range of the projectile launched upwards from the same point at an angle of 45degree to the horizontal with the same speed is??

OpenStudy (anonymous):

options are 1.5km 3km 6 km 4.5 km

OpenStudy (anonymous):

the answer is 3 km

OpenStudy (anonymous):

substitute the given parameters in the equation for range. that will give you the value of initial velocity squared. now substitute the parameters again but with theta = 45. you will get the answer

OpenStudy (anonymous):

3 km [range = (v ^2\times \sin 2\theta)/2g\]

OpenStudy (anonymous):

ok i get it...

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