Answer:
All the competitors will move with the same velocity.
Explanation:
Here, the situations for each competitor are identical. Thus, they will exert the same force and hence, their velocities at each instants will be identical.
Force exerted by the bullet = mass * acceleration = 0.013 * 850 = 11.05 Newtons.
the rifle exerts same force in opposite direction so we have
11.05 = 3.5 * a
acceleration = 11.05 / 3.5 = 3.16 m /s^-2
Answer:
0.8726 
Explanation:
We are to convert 1.85 x
to 
First, let us convert the numerator from ft3 to m3
1 ft3 = 0.0283 m3
Hence,
1.85 x
ft3 = 1.85 x
x 0.0283 m3
= 52.355 m3
Now, let us convert the denominator from minutes to seconds
1 min = 60 sec
Therefore;
1.85 x
= 52.355/60 
= 0.8726 
Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ =
/ vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m