If you can swim in still water at 0.5m/s, the shortest time it would take you to swim from bank to bank across a 20m wide river,
if the water flows downstream at a rate of 1.5m/s, is most nearly:
2 answers:
Answer:42.55 seconds
Explanation:
Check attached file for the diagram.
Vr= velocity of the river, if the water flows downstream, which is equal to 1.5m/s(from the question), Vb= my velocity at angle alpha which is perpendicular to the x-axis, and c= distance to swim from river bank to the other river bank.
Therefore, the shortest time to be taken to swim or cross the river,t=river with width of C÷ vertical component Vj.
Note that (vector) Vj= (vector)Vr+(vector)Vb.
Also, alpha must be maximum and for that to happen it must have a value of one(1).
Hence, time taken to cross the river,t= C/Vb× sin (alpha).
t= 20m÷ 0.5 ×sin 90.
t= 20÷ 0.47.
t= 42.55 seconds.
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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°
Refer to the diagram shown below.
Assume that air resistance is ignored.
Note:
The distance, h, of a falling object with initial vertical velocity of zero at time t is
h = (1/2)gt²
where
g = 9.8 m/s²
The initial vertical velocity of the supplies is 0 m/s.
It the time taken for the supplies to reach the ground is t, then
(50 m) = (1/2)*(9.8 m/s²)*(t s)²
Hence obtain
t² = 50/4.9 = 10.2041
t = 3.1944 s
The horizontal distance traveled at a speed of 100 m/s is
d = (100 m/s)*(3.1944 s) = 319.44 m
Answer: 319.4 m (nearest tenth)
Assume that air resistance is ignored.
Note:
The distance, h, of a falling object with initial vertical velocity of zero at time t is
h = (1/2)gt²
where
g = 9.8 m/s²
The initial vertical velocity of the supplies is 0 m/s.
It the time taken for the supplies to reach the ground is t, then
(50 m) = (1/2)*(9.8 m/s²)*(t s)²
Hence obtain
t² = 50/4.9 = 10.2041
t = 3.1944 s
The horizontal distance traveled at a speed of 100 m/s is
d = (100 m/s)*(3.1944 s) = 319.44 m
Answer: 319.4 m (nearest tenth)