Question

An archer stands 1.3 m tall and fires an arrow from this height at a target 200 m away. The centre of the target is 1.8 m above the ground and the arrow is fired at an angle of a to the horizontal with an initial speed of 60 m/s. i. Sketch a diagram to illustrate the information above and label the necessary components. ii. Calculate the two possible angles, a, that would allow the archer to hit the target. iii. Out of the two possible angles found above, state which angle the archer should choose and explain why.

Answer 1

Vosind obom/s Dod Volosa Y= 1.31 92 1.8m 2oom 1) nooizontal equation is nex. + dorth at 2 Ho initial point = 0 Vox= horizontal component de of No = Volosa time in the air t a = acceleration in the a direction=0 in n = v. Cosat & Cosa 3:34 - 0 Cosa a 200 bolosa t air

Now in the direction, У 9 = 7 + Voyt & I gt2 – 1 Jo = 1.3m , y = 1.8m f=tair Voy Nosind = 60 sind G = 9.8 m/s2 negative because motion is upward From Requly 108 1.3 + 6o sind 13.34 3497.5) ( 3.3412 Cosa 54.66 => 0.5 = 200.4 tand 60524 0.5 Cos22 & 200.4 taux x cost – 54.66 => 0. 5 (0324 +54.66 = 200.4 (sind losa) -

-> 2 0.5 (1-sin22) + 54.66 = 200.4 sin 22 => 0.5-0.5 sind +54.66 [:2 Sinalosa Sin 20 = 100.2 Sinza - 100.2 Sinaa + 0.5 Sinza 55.16 = 0 => 100.2 Sin 22 +0,5 (1 - Los 22) - 55.16=0 2 [! sin2=(1-(0322) 2 => 100. 2 Sinza 0.25 los 22 -54.91= 0 3 We can write A sing + B Coso = R Cos (0-0) where, R=SA2+B2 & 0 = tans (2) here A= 100. 2 B = -0.25, 0 = 24 ) ..R= 100.2 6 / Ø = - 89.85 from equ. 100. 2 Cos (24 + 89.85) = 54.9) => Cos (22 + 89.85) 20 + 89,85 = 65'(0.548)=656.76) = 0.548 =) 2) 2 4 - - 33.09. 712= de-16.54 this angle is negative But we can take it positive los (0) = cos(-o) So L = 16.54 dry

his) another angle is [x = -16.54 + 2x , as Cos (0) = cos(2466 => = 343.46 x = 16.54 as the height is upward not downward so & should be positive and between oo to 27 and another 2 not possible. so,