Calculate how fast the ball would be moving at the instant it leaves the projectile launcher of the spring is compressed by 3.75
1 answer:
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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°