Question

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?

Answer 1

Answer:

Write equation 1 for motion in the x-direction:

Sx=(Vi)xt+12axt2

ax=0 because we assume negligible air resistance:

Sx=(Vi)xt

75=(35)cosθt —— equation 4

2 unknowns in this equation so we need another equation to solve it . . .

Now write equation 1 for motion in the y-direction:

Sy=(Vi)yt+12ayt2

I will assume positive= up. Also, Sy=0 since the ball returns to the same vertical position.

0=(35sinθ)t+12(−9.81)t2

cancel “t” in all terms gives

0=35sinθ+12(−9.81)t

or

t=709.81sinθ

substitute this into equation 4 gives:

75=35cosθ[709.81sinθ]

or

75=24509.81sinθcosθ

but sinθcosθ = sin2θ2

75=24509.81sin2θ2

θ=18.5∘

Answer 2

Given: 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. Its initial speed is 37m/s.

To find: Angle at which the arrow must be thrown

Solution: The arrow is shot at some angle a. This is a case of oblique projectile motion. Since the target is at same height as release of the arrow, therefore range of projectile= 75 m

u = 37 m/s

The formula for range in a projectile motion is given by:

$r = \frac{ {u}^{2} \sin(2a) }{g}$

$= > 75 = \frac{ {(37)}^{2} \sin(2a) }{10}$

=> sin(2a) = 75 × 10 / 1369

=> sin 2a = 0.55

=> sin 2a = sin 34° ( sin 34° = 0.55)

=> 2a = 34°

=> a = 34/2

=> a = 17°

Now, since range is same for angles a and 90-a, therefore, at (90-17)= 73°, the range is the same.

Therefore, the angle at which the arrow must be thrown is 17° or 73°.