The distance of the length YD is 3.57m, the velocity of the object at point D is 27m/s and the time required for the object to reach the ground is 0.17s
Data given;
a)
[tex]YD=CDsin[/tex]θ = [tex]5sin45=\frac{5}{\sqrt{2} } = 3.57m[/tex]
b)
The change in kinetic energy is given as
Δ in kinetic energy = Δ in potential energy + work done by friction
K.E - 1/2m[tex]v^2[/tex] = mgh[tex]_1[/tex] - mgh[tex]_2[/tex] + (-μmg.x)
K.E = mg(50 - 3.57) + (-mg(0.3*100) + 1/2 m[tex]v^2[/tex]
[tex]\frac{1}{2}mv^2_f=mg(46.43)-mg(30)+\frac{1}{2} m(400)\\\\v^2_f=20(16.43)+400\\v^2_f=728.6\\v=\sqrt{728.6} \\v_f=26.99 = 27m/s[/tex]
The velocity of the object at D with a distance of 5m.
c)
The velocity of the object in the y-axis is
[tex]v_y=vsin45=19.09[/tex]
Acceleration in y-axis = 9.8
Height = 3.57m
h = [tex]ut+\frac{1}{2}at^2[/tex]
[tex]3.57=19.28(t)+\frac{1}{2}(9.8)t^2\\3.57=19.28t+4.9t^2\\4.9t^2+19.28t-3.57=0\\a=4.9, b=19.28, c= -3.57\\t=\frac{-19.28+\sqrt{(19.28)^2-4(4.9)(-3.57)} }{(2*4.9)} \\t=\frac{-19.28+21.02}{9.8}[/tex]
Taking the positive value
[tex]\frac{-19.28+21.02}{9.8}=0.17s[/tex]
The time required for the object to reach ground is 0.17s
From the calculations above, the distance YD is 3.57m, the velocity of the object at point D is 27m/s and the time required for the object to reach ground is 0.17s
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