Question

The highest barrier that a projectile can clear is 12.7 m, when the projectile is launched at an angle 19.0° above the horizontal. What is the projectile's launch speed?

Answer 1

Step-by-step explanation:

h=((usinx)^2)/2g

12.7=((usin(19))^2)/(2×9.8)

(12.7×2×9.8)=(usin(19))^2

248.92=(usin(19))^2

usin19 =√248.92 =15.78

U =15.78/(sin19)

=48.46m/s----> speed