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Based on the data collected, explain why a launch angle of 30 degrees (HIGH HORIZONTAL VELOCITY & SMALL TIME OF FLIGHT) will allow a projectile to go just as far when the launch angle of 60 degrees (LOW HORIZONTAL VELOCITY AND LARGE TIME OF FLIGHT).

User Luke Vo
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Answer:

Step-by-step explanation:

The horizontal distance traveled by the projectile is given by the formula


R=(u^(2)Sin2\theta )/(g)

The formula for the time of flight is given by


T = (2u Sin\theta )/(g)

Case I: when the launch angle is 30°

So,
R_(1)=(u^(2)Sin60 )/(g)


R_(1)=(0.866u^(2))/(g)

Horizontal velocity = u Cos 30 = 0.866 u


T_(1) = (2u Sin30 )/(g)=(u)/(g)

Case II: when the launch angle is 60°


R_(2)=(u^(2)Sin120)/(g)


R_(2)=(0.866u^(2))/(g)

Horizontal velocity = u Cos 60 = 0.5 u


T_(1) = (2u Sin60 )/(g)=(1.73 u)/(g)

By observing the case I and case II, we conclude that

R1 = R2

Horizontal velocity 1 > Horizontal velocity 2

T1 < T2

User Ujjawal Bhandari
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