Let at any instant of time the speed is vo and the angle made by the bike with the horizontal is given
now we have
component of speed in x direction given as
component of speed in y direction will be
now from above two equations we can say that here
= angle with the horizontal at any instant
and since here it is a sine curve so we know that
so we have slope of graph
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°
Answer:
The third particle should be at 0.0743 m from the origin on the negative x-axis.
Explanation:
Let's assume that the third charge is on the negative x-axis. So we have:
We know that the electric field is:
Where:
So, we have:
Let's solve it for r(3).
Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.
I hope it helps you!