26. The horizontal range of a projectile thrown at a angle (alpha) with the horizontal is half of its maximum range, when thrown at same speed. Here alpha is equal to

60 degree

45 degree

30 degree

15 degree

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Solution

Dear Student,

$m a x i m u m r a n g e i s a t 45 d e g r e e p r o j e c t i o R_{m a x} = \frac{u^{2} \sin (2 * 45)}{g} = \frac{u^{2}}{g} a t a l p h a t h e r a n g e b e c o m e s R_{m a x} / 2 \frac{R_{m a x}}{2} = \frac{u^{2} \sin (2 * α)}{g} \frac{1}{2} = \sin (2 * α) 2 α = 30 d e g r e e α = 15 d e g r e e R e g a r d s$

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26. The horizontal range of a projectile thrown at a angle (alpha) with the horizontal is half of its maximum range, when thrown at same speed. Here alpha is equal to 60 degree 45 degree 30 degree 15 degree

Dear Student, maximum range is at 45 degree projectio Rmax=u2sin2*45g=u2gat alpha the range becomes Rmax/2Rmax2=u2sin2*αg12=sin2*α2α=30 degreeα=15 degreeRegards

26. The horizontal range of a projectile thrown at a angle (alpha) with the horizontal is half of its maximum range, when thrown at same speed. Here alpha is equal to

60 degree

45 degree

30 degree

15 degree