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Edit View History Bookmarks People Window Help rse Dashboard x , D created by Camtasia x : Solved: The position o, x fi MyUW ot Secure ezto.mheducation.com/hm.tp alue 0.00 points 1 out of 3 attempts Do not round intermediate calculations, however for display purposes report intermediate steps rounded to four significant figures. Give your final answer(s) to three significant figures. In a movie scene involving a car chase, a car goes over the top of a ramp at A and lands at B below. Determine the speed of the car at4 if the car is to cover distance d -160 ft for a -23 and B-270. Neglect aerodynamic effects. v o ft/s References eBook & Resources Multipart Answer Chapter: 12-Particle Kinematics Section: 01 Chapter Problems

Explanation / Answer

In this type of projectile motion there will be acceleration in both vertical and horizontal direction.

Now acceleration in xy-coordinate system will be (where x-axis is AB and y-axis is perpendicular to AB)

alpha = A & beta = B

a = g*sin B i - g*cos B j

So, ax = g*sin B

ay = -g*cos B

Now Given that intial velocity is V0 at angle of (A + B) deg above the horizontal, So

V0x = V0*cos (A + B)

V0y = V0*sin (A + B)

Now Using 2nd kinematic equation in y-direction

y = V0y*t + 0.5*ay*t^2

y = V0*sin (A + B)*t - 0.5*g*cos B*t^2

We can see that from point A to point B, when car land, vertical displacement is zero. So y = 0

0 = V0*sin (A + B)*t - 0.5*g*cos B*t^2

V0*sin (A + B)*t = 0.5*g*cos B*t^2

t = [2*V0*sin (A + B)]/(g*cos B)

Now Horizontal distance covered in x-direction will be

Using 2nd kinematic equation inx-direction

d = V0x*t + 0.5*ax*t^2

d = V0*cos (A + B)*t + 0.5*g*sin B*t^2

Using the value of t from first bold equation

d = V0*cos (A + B)*[2*V0*sin (A + B)]/(g*cos B) + 0.5*g*sin B*[[2*V0*sin (A + B)]/(g*cos B)]^2

Now in this we have every value given in question, other that V0, So

Rearranging above equation

V0^2 = d/[2*cos (A + B)*sin (A + B)/(g*cos B) + 2*g*sin B*[[sin (A + B)]/(g*cos B)]^2]

Now given that

d = 160 ft

A = 23 deg

B = 27 deg

A + B = 50 deg

g = 32.2 ft/sec^2

Using these values

V0^2 = 160/[2*cos 50 deg*sin 50 deg/(32.2*cos 27 deg) + 2*32.2*sin 27 deg*[[sin 50 deg]/(32.2*cos 27 deg)]^2]

V0^2 = 2900.19

V0 = sqrt (2900.19)

V0 = 53.85 ft/sec

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Comment below if you have any doubt in any step.

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