Robin would like to shoot an orange in a tree with his bow and arrow. the orange is hanging yf=5.00 m above the ground. on his first try, robin looses the arrow at =36.0 m/s at an angle of theta=30.0° above the horizontal. the arrow has an initial height of y0=1.30 m, and its tip is x=49.0 m away from the target orange. treating the arrow as a point projectile and neglecting air resistance, what is the height of the arrow once it has reached the horizontal position x of the orange? use =9.81 m/s2 for the acceleration due to gravity.

17.5 m

Explanation:

First of all, we need to find the time the arrow need to cover the horizontal distance between the starting point and the orange, which is

x = 49.0 m

We start by calculating the horizontal component of the arrow's velocity:

And this horizontal velocity is constant during the entire motion. So, the time taken to reach the horizontal position of the orange is

Now we can find the height of the arrow at that time by using the equation for the vertical position:

where:

h = 1.30 m is the initial height

is the initial vertical velocity

t = 1.57 s is the time

is the acceleration of gravity

Substituting into the equation, we find

anchor = archer //

not too sure thinking all accelerations are pointed down