Verified answer

y = sin (27)Vo - 1/2 at^2

x = Vot

y=0 @ target so 4.9t^2 = sin(27)Vo and t^2 = 10 sin(27)

so t = 2.13 s

x = 49 * 2.13 = 104.4 m

Use kinematic equations: v = v0 + at x = v0t + .5at^2 v^2 - v0^2 = 2ax For area A: you may opt to locate the optimal suitable, so stumble on the vertical velocity ingredient and plug into the equation ----> v^2 - v0^2 = 2ax v0 = 40 9 sin(30) v = 0 (by actuality at the same time as the arrow reached max suitable, it extremely is 0) a = 9.80 one x = the unknown. 0^2 - 24.5^2 = 2*9.80 one*x x = 30.59m severe area B utilising the first answer, freshen up for entire time interior the air, then used the price*time = distance formula to freshen up for the area. Use equation a million to freshen up for time, t = 2.497s, yet multiply by 2 by actuality the time it takes to get from 0 to 30.59m is two.497 s, and the time it takes to come back off is two.497s so the finished time is 4.99s. stumble on the horizontal ingredient of the price to locate the area. 49cos(30) = 40 2.40 4 distance = velocity*time distance = 40 2.40 4*4.ninety 9 d = 2111.75m