1. Ch-4
Help-Session

2. CH-4-072
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
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
4/25/2019

3. 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)
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.
Angle between vbw and vbg=90-60=30°
vbw=vbg cos 30=0.4 cos 30=0.346 m/s
vwg
vbw
vbg
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
4/25/2019

4. CH-4-071 T=(2πR)/v=(2π*π)/4 =(π*π)/2
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* *9.8*4.3*4.3
= = m
h = 264 m
4/25/2019

5. 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
vAG
vBG =30 cos 60 i-30sin60j
=15i-26j
vAG= -40 cos 30i-40 sin 30j
=-34.6i-20j
vAG - vBG= ( )i-(20-26)j
=-49.6 I +6j
|vAG - vBG |= = m/s
4/25/2019