Listen if you have to cheat for the dba thing not worth even doing that dba tbh this is very easy as I just did it in like 5 minutes it gives you everything you need even the formulas so use your f .u .c .k. 1 .n g brain you monkey...
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
I think it may be c i learned about this last year
Answer:
7.6 s
Explanation:
Considering kinematics formula for final velocity as

Where v and u are final and initial velocities, a is acceleration and s is distance moved.
Making v the subject then

Substituting 8.8 m/s for u, 138 m for s and 2.45 m/s2 for a then

Also, v=u+at and making t the subject of the formula

Substituting 27.45 m/s for v, 8.8 m/s for u and 2.45 m/s for a then

Therefore, it needs 7.6 seconds to travel
Wave C
because it is the closest or each wave is closer than the others