Answer:
the shooting angle ia 18.4º
Explanation:
For resolution of this exercise we use projectile launch expressions, let's see the scope
       R = Vo² sin (2θ) / g
       sin 2θ = g R / Vo²
       sin 2θ = 9.8 75/35²
       2θ = sin⁻¹ (0.6)
       θ = 18.4º
To know how for the arrow the tree branch we calculate the height of the arrow at this point
        X2 = 75/2 = 37.5 m
We calculate the time to reach this point since the speed is constant on the X axis
        X = Vox t
        t2 = X2 / Vox = X2 / (Vo cosθ)
         t2 = 37.5 / (35 cos 18.4)
         t2 = 1.13 s
With this time we calculate the height at this point
         Y = Voy t - ½ g t²
         Y = 35 sin 18.4   1.13 - ½ 9.8 1,13²
         Y = 6.23 m
With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch
 
        
                    
             
        
        
        
When looking for distance you multiply speed by time
So 15 x 2 = 30
30 is the distance between his house and school
        
             
        
        
        
Answer:
The actual angle is 30°
Explanation:
<h2>Equation of projectile:</h2><h2>y axis:</h2>

the velocity is Zero when the projectile reach in the maximum altitude:

When the time is vo/g the projectile are in the middle of the range.
<h2>x axis:</h2>

R=Range


**sin(2A)=2sin(A)cos(A)
<h2>The maximum range occurs when A=45°
(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>
Let B the actual angle of projectile

2B=60°
B=30°