1w 4 Begi Date 919 2018 12:0 :00 AM Due Date 9/27/201 81 1:59:00 PM End Date (S%

Question

1w 4 Begi Date 919 2018 12:0 :00 AM Due Date 9/27/201 81 1:59:00 PM End Date (S%) Problem i0: A dart gun contains a spring of spring constant k 16 N/m which is used to fire a dart of mass m. The dart leaves the gun at a speed ofv=4.5 m/s after the spring is compressed 1 cm. 12/16 201 81159 00PM Ct t, 33% Part (a) What is the weight, F, in Newtons, of the dart? 7.74 104 -7.74E-4 Correct 33% Part (b) what is the dart's speed when it hits the floor ve in ms, ift is fired horizontally at a height of h 2 meters? -771 Correct! ** 33% Part (c) what angle, in degrees, does the dart's final velocity make with the horizontal? Potential 98% 4 5 6 Attempts remaining cotan asin (Cls per attempt) detailed view acotan cosh( tanh0 cotanh0

Explanation / Answer

If you need any help with part A and B comment below.

Part C.

We know that in projectile motion horizontal velocity remains constant, So final horizontal velocity will be

Vfx = V0x = 4.5 m/sec

final vertical speed will be using 3rd kinematic equation

Vfy^2 = Viy^2 + 2*a*d

Viy = 0 m/sec, since initiallly only horizontal velocity

a = -g = -9.81 m/sec

h = -2 m

Vfy = sqrt (0^2 + 2*9.81*2)

Vfy = -6.26 m/sec (-ve sign because velocity is downward)

Now

Vf = sqrt (Vfx^2 + Vfy^2)

Vf = sqrt (4.5^2 + (-6.26)^2)

Vf = 7.71 m/sec

Now direction will be given by:

Direction = arctan (Vfy/Vfx)

= arctan (6.26/4.5)

= 54.3 deg below the horizontal

(Since angle is below the horizontal so direction will be negative)

Direction = -54.3 deg (See that if negative answer does not work try the magnitude of angle.)

Please Upvote.

Comment below if you have any doubt.

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