(8c8p76) A 0.450 ko projectile is launched from the edge of a cliff with an init

Question

(8c8p76) A 0.450 ko projectile is launched from the edge of a cliff with an initial kinetic energy of 1650J and at its highest point is 135.5 above the launch point. What is the horizontal cormponent of its velocity? Subrmil ArISwEl Tries 0/3 What was the vertical component of its velocity just alter laun? Submit Answer Tries 0/s At one instant during its flight the vertical component of its velocity is 10.25 m/s. At that time, how far is it above or below the launch point? Submit Ariswer Tries 0/8

Explanation / Answer

a) Apply conservation of energy

KEf + m*gh = KEi

(1/2)*m*vx^2 + m*g*h = KEi (at heighest point y-component of velocity vanishes)

(1/2)*0.45*vx^2 + 0.45*9.8*135.5 = 1650

==> vx = 68.4 m/s

b) KEi = (1/2)*m*vi^2

vi = sqrt(2*KEi/m)

= sqrt(2*1650/0.45)

= 85.6 m/s

we know, vi = sqrt(vx^2 +vy^2)

vi^2 = vx^2 + vy^2

==> vy^2 = vi^2 - vx^2

==> vy = sqrt(vi^2 - vx^2)

= sqrt(85.6^2 - 68.4^2)

= 51.5 m/s

c) y = (v2^2 - v1^2)/(2*g)

= (40.25^2 - 51.5^2)/(2*(-9.8))

= 52.7 m

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