(8c8p76) A 0.700 kg projectile is launched from the edge of a cliff with an init
ID: 1389012 • Letter: #
Question
(8c8p76) A 0.700 kg projectile is launched from the edge of a cliff with an initial kinetic energy of 1625 J and at its highest point is 110.2 in above the launch point. What is the horizontal component of its velocity? Submit Answer Tries 0/5 What was the vertical component of its velocity just after launch? Submit Answer Tries 0/5 At one instant during its flight the vertical component of its velocity is 29.02 m/s. At that time, how far is it above or below the launch point? Submit Answer Tries 0/5Explanation / Answer
a) let vox and voy are horizonatl compoents of initial velocity.
Apply enrgy conservation
Initail mechanical enrgy = final mechanical enrgy
1625 = m*g*h + 0.5*m*vox^2
1625 = 0.7*9.8*110.2 + 0.5*0.7*vox^2
1625 = 756 + 0.5*0.7*vox^2
1625 - 756 = 0.5*0.7*vox^2
869 = 0.5*0.7*vox^2
==> vox = sqrt(2*869/0.7)
= 49.8 m/s <<<<<<<<<<<<-------------Answer
b)H = voy^2/(2*g)
voy = sqrt(2*g*H)
= sqrt(2*9.8*110.2)
= 46.5 m/s <<<<<<<<<<<<-------------Answer
c) Apply, vy^2 - voy^2 = -2*g*h
==> h = (voy^2 - vy^2)/(2*g)
= (46.5^2 - 29.02^2)/(2*9.8)
= 67.4 m above the cliff <<<<<<<<<<<<-------------Answer
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