Question

An object with mass of 4kg is thrown with initial velocity of 20m/s from point A and follows the track of ABCD.

Answer 1

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;

• Mass = 4kg

• Initial velocity = 20m/s

• length CD = 5m (from the image given)

a)

Determine The Length of YD

$YD=CDsin$θ = $5sin45=\frac{5}{\sqrt{2} } = 3.57m$

b)

The velocity of the object at point D

The change in kinetic energy is given as

Δ in kinetic energy = Δ in potential energy + work done by friction

K.E - 1/2m$v^2$ = mgh$_1$ - mgh$_2$ + (-μmg.x)

K.E = mg(50 - 3.57) + (-mg(0.3*100) + 1/2 m$v^2$

$\frac{1}{2}mv^2_f=mg(46.43)-mg(30)+\frac{1}{2} m(400)\\\\v^2_f=20(16.43)+400\\v^2_f=728.6\\v=\sqrt{728.6} \\v_f=26.99 = 27m/s$

The velocity of the object at D with a distance of 5m.

c)

The the required for the object to reach ground

The velocity of the object in the y-axis is

$v_y=vsin45=19.09$

Acceleration in y-axis = 9.8

Height = 3.57m

h = $ut+\frac{1}{2}at^2$

$3.57=19.28(t)+\frac{1}{2}(9.8)t^2\\3.57=19.28t+4.9t^2\\4.9t^2+19.28t-3.57=0\\a=4.9, b=19.28, c= -3.57\\t=\frac{-19.28+\sqrt{(19.28)^2-4(4.9)(-3.57)} }{(2*4.9)} \\t=\frac{-19.28+21.02}{9.8}$

Taking the positive value

$\frac{-19.28+21.02}{9.8}=0.17s$

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

Learn more on projectile motion here;

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