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Ch-4. Help-Session. Q9: A projectile is fired over a flat horizontal land. It takes 10 s to reach its range of 100 m. What is the speed of the projectile at the highest point of its trajectory?.

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## Ch-4

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**Ch-4**Help-Session**Q9: A projectile is fired over a flat horizontal land. It**takes 10 s to reach its range of 100 m. What is the speed of the projectile at the highest point of its trajectory? Q10.: A particle is moving counterclockwise in x-y plane in a uniform circular motion. The circle is centered at the origin and has a radius of 2.0 m. When the velocity of the particle is 4i m/s , then its acceleration is CH-4-072 • x1=vit+at2/2 |a|=|v|2 /R=16/2=8 m/s2 Particle is in at beginning of fourth Quadrant of circle, a towards center i.e. j a= (8 m/s2) j At maximum height h v= v0xi+vyj; vy=0 v0x= R/t=100/10=10 m/s v= v0xi = (10 m/s )i**Q11.: A river is flowing 0.20 m/s east. A boat in this river**has a speed of 0.40 m/s directed 60° south of east relative to the earth. Find the velocity of the boat relative to the river. vbg vbw vwg CH-4-072 Q12.: A particle has its position vector defined by R=(2.ot-t2)i+(3.0t-1.5t2)j. At what time is its speed equal to zero? (Ans: 1.0 s) • Angle between vbw and vbg=90-60=30° • vbw=vbg cos 30=0.4 cos 30=0.346 m/s v=dr/dt=(2-2t)i+(3-3t)j To calculate t for v=0 Equate quotient of i and j separately to Zero we get 2-2t=0; t=1 3-3t=0; t=1**CH-4-071**T071 :Q10.A certain airplane has a speed of 80.6 m/s and is diving at an angle of 30.0° below the horizontal when it releases an object. The horizontal distance from the point of release was 300 m as shown in Fig.4. How high was the point of release of the object? (Ans: 264 m) Q11.: An object is moving on a circular path of radius π meters at a constant speed of 4.0 m/s. The time required for one revolution is: (Ans: (π2/2) s) T=(2πR)/v=(2π*π)/4 =(π*π)/2 Time t for x-motion= 300/(80.6*cos30) =4.30 s -h=-80.6*sin30*4.3-0.5*9.8*4.3*4.3 =-173.3-90.6=-263.9 m h = 264 m**vBG**vAG CH-4-071 Q12.: Ship A travels 40 km/h in a direction of 30° West of North and ship B travels 60° East of North at 30 km/h. What is the magnitude of the velocity of ship A relative to ship B?( Ans: 50 km/h) vBG =30 cos 60 i-30sin60j =15i-26j vAG= -40 cos 30i-40 sin 30j =-34.6i-20j vAG - vBG= (-34.6-15)i-(20-26)j =-49.6 I +6j |vAG - vBG |=49.62+62= 49.96 m/s

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