You would like to shoot an orange in a tree with your bow an

You would like to shoot an orange in a tree with your bow and arrow. The orange is hanging 5.00 m above the ground. On your first try. You fire the arrow at 31.0 m/s at an angle of 30.0 degree above the horizontal from a height of 1.20 m while standing 54.0 m away. Treating the arrow once it has traveled the 54.0 m horizontally to the tree?

Solution

initial velocity of the projectile v = 31.0 m/s

angle of projection = 30⁰

vertical component of the velocity v_y = 31.0 Sin(30) = 15.5 m/s

Horizontal component v_x = 31.0Cos(30) = 26.85 m/s

vertical position of the arrow at any time t is given by

y(t) = v_y t-gt²/2 = 15.5t-9.8t²/2

let us consider the ground point at the start as the origin then the position of

y(t) = 1.2 +15.5t-9.8t²/2 -(1)

Horizontal position is given by

x(t) = v_xt = 26.85t ----------(2)

and (2) define the position the arrow at any time with respect to our defined origin on the ground at the start.

when the arrow traveled 54.0 m horizontally

54.0 = 26.85t or t= 54.0/26.85 = 2.01

substituting t = 2.01 in (1) we get the y position

y(2.01) = 1.2 +15.5*2.01 -9.8*(2.01)²/2

             = 12.56 m

to reach the arrow y(t) = 5.0

5=1.2 +15.5t-9.8t²/2

4.9t² -15.5t +3.8 = 0

solving the above quadratic for t

we get t = 0.27, 2.89

substituting for t in (2)

x= 7.25 m , 77.6m

the projectile goes up reaches a maximum height and returns to ground thus taking a curved path, the two values of t and y indicate the two point on the path where it can meet the target, first value is in the ascending and second value while descending.

x= 77.6 the choice.