An archer shoots an arrow

An archer shoots an arrow at a 75 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. If d = 75 m and v = 37 m/s. At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 37m/s?

Solutions

Expert Solution

Step 1

$W e k n o w t h a t, T h e e q u a t i o n f o r m o t i o n i n x - d i r e c t i o n, S x = u x t + 12 a x t 2 a x = 0 b e c a u s e w e a s s u m e n e g l i g i b l e a i r r e s i s tan c e . S o, S x = u x t 75 = 37 cos θ t . . . . . . . . . . . . . . . . . . . . . (1) T h e r e a r e t w o u n k n o w n s i n t h i s e q u a t i o n s o w e n e e d a n o t h e r e q u a t i o n t o s o l v e i t . N o w w r i t e t h e e q u a t i o n f o r m o t i o n i n t h e y - d i r e c t i o n; S y = u y t + 12 a y t 2 S y = 0, sin c e t h e b a l l r e t u r n s t o t h e s a m e v e r t i c a l p o s i t i o n . S o, 0 = 37 sin θ t + 12 - 9.81 t 2 37 sin θ - 4.905 t = 0 . . . . . . . . . . . . . . . . . . . . . . (2) "> W e k n o w t h a t, T h e e q u a t i o n f o r m o t i o n i n x - d i r e c t i o n, S x = u x t + 1 2 a x t 2 a x = 0 b e c a u s e w e a s s u m e n e g l i g i b l e a i r r e s i s tan c e . S o, S x = u x t 75 = 37 cos (θ) t . . . . . . . . . . . . . . . . . . . . . (1) T h e r e a r e t w o u n k n o w n s i n t h i s e q u a t i o n s o w e n e e d a n o t h e r e q u a t i o n t o s o l v e i t . N o w w r i t e t h e e q u a t i o n f o r m o t i o n i n t h e y - d i r e c t i o n; S y = u y t + 1 2 a y t 2 S y = 0, sin c e t h e b a l l r e t u r n s t o t h e s a m e v e r t i c a l p o s i t i o n . S o, 0 = 37 sin (θ) t + 1 2 (- 9.81) t 2 37 sin (θ) - 4.905 t = 0 . . . . . . . . . . . . . . . . . . . . . . (2)$

Step 2

$F r o m e q u a t i o n (1) a n d (2), w e g e t 75 = 37 cos θ \times 37 sin θ 4.905 sin 2 θ = 0.537 θ = 16 . 25 o A n d, t = 37 sin 16 . 25 o 4.905 t = 2.11 s e c "> F r o m e q u a t i o n (1) a n d (2), w e g e t 75 = 37 cos (θ) \times 37 sin ( θ ) 4 . 905 sin (2 θ) = 0.537 θ = 16 . 25 o A n d, t = 37 sin ( 16 . 25 o ) 4 . 905 t = 2.11 s e c$

Raffay answered 1 year ago

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