Homework # 3 (Curvilinear Motion) 1. A particle is fire at 10 m/s from a incline

Question

Homework # 3 (Curvilinear Motion) 1. A particle is fire at 10 m/s from a incline. The angle from the incline surface and the launched vector is 90°. Determine de range R. 0 m/s 2. A wind-tunnel analysis indicate that a car starting from Point A at rest was a tangential acceleration of a 0.4-.001 v2, where v is the car velocity in m/s. Determine the velocity and acceleration of the car when reaches B? 200 m 3. The coordinates of point A of the crane are given as function of time in seconds by r 4+0.3t [m and 0 0.04 t [rad]. Determine the acceleration of point A at 3 seconds. 4. The quarterback throws the football with velocity Vo [ft/s Jat 45o above the horizontal. At the same instant, the receiver standing 20 ft in front of him start running straight downfield at 10 [ft/s] and catches the ball. Assume that the ball is thrown and caught at the same height above the ground. What is the velocity Vo? 5. A broad jumper approach his takeoff board A with a horizontal velocity of 30ft/s. Determine the vertical component vy of the velocity of his center of gravity at takeoff for him to make the jump shown. What is the vertical rise h of his center of gravity?

Explanation / Answer

Q1- The range of projectile down the inclined plane is given by the formula-

R= (u^2/g(cosa)^2)(sin(2b+a)+sina)

Here, a is the angle of inclination of plane and b is the angle of projection with horizontal.

Now here given, b=alpha-a=90°-30°=60° and a=30° and u=initial velocity of projection= 10 m/s.

So put all the values in above expression.

R=[10^2/(9.81×(cos30)^2)][sin(2×60+30)+sin(30)]

So, R=[100/(9.81×(3/4))][sin(150)+sin(30)]

Now we know sin150=sin(90+60)=cos60 put above,

R=[400/(9.81×3)][cos60+sin30]

R=[400/29.43][1/2+1/2]

So, R=400/29.43=13.6 m answer.

Please ask if you have any doubt further in comment box.

Thanks.

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