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The highest barrier that a projectile can clear is 12.2 m, when the projectile is launched at an angle of 13.0 ° above the horizontal. what is the projectile's launch speed?

Respuesta :

consider the motion in vertical direction

v = initial velocity in the vertical direction = ?

v' = final velocity in the vertical direction at the highest point = 0 m/s

a = acceleration due to gravity = - 9.8 m/s²

h = maximum height = 12.2 m

Using the kinematics equation

v'² = v² + 2 a h

0² = v² + 2 (- 9.8) (12.2)

v = 15.5 m/s

let assume that velocity of launch be v₀

θ = angle of launch = 13

vertical component of the velocity of launch is given as

v = v₀ Sinθ

15.5 = v₀ Sin13

v₀ = 69 m/s

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