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
Answer:
2.5m/s^2
Explanation:
Step one:
given
distance = 20meters
time = 2 seconds
initial velocity u= 0m/s
let us solve for the final velocity
velocity = distance/time
velocity= 20/2
velocity= 10m/s
divide both sides by 40
the awnser to ur question is D
Answer:
Final velocity v=19.83 m/sec
Explanation:
We have given initial velocity u =5.13 m/sexc
Acceleration of automobile 
Time t =4.9 sec
We have to find the final velocity v
According to first law of motion v = u+at ,here v is the final velocity , a is acceleration and t is time
So 
So the final velocity is 19.83 m/sec
