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# Velocity of horizontal motion is constant. So;

Kinematics Exam2 and Problem Solutions

1. An object is dropped from 320 m high. Find the time of motion and velocity when it hits the ground.(g=10m/s2)

h=1/2.g.t2 , v=g.t

h=320m

g=10m/s2

320=1/2.10.t2

t=8s.

v=g.t=10.8=80m/s

2.An object does free fall and it takes 60m distance during last 2 seconds of its motion. Find the height it is dropped.(g=10m/s2) t is the time of motion

h=1/2.g.t2

h1=1/2.g.t12

put t1=t-2 and h-h1=60 in the equation,

1/2.g.t2-1/2.g.t12=60

5t2-5(t2-4t+4)=60

t=4s

h=1/2.g.t2=1/2.10.42=80m

3. An object is dropped from 144m height and it does free fall motion. Distance it travels and time of motion are given in the picture below. Find the distance between points B-C. We can draw velocity time graph of object and area under this graph gives us position of the object. As you can see from the velocity time graph, object travels 5h distance during 2t-3t which is the distance between the points B and C.

All distance traveled is 36h

144m=36h

h=4m

Distance between B-C=5h=5.4m=20m

4. A hot- air balloon having initial velocity v0 rises. Stone dropped from this balloon, when it is 135 m height, hits the ground after 9 s. Find the velocity of the balloon. -h=v0.tfligth-1/2.g.tfligth2

-135=v0.9-1/2.10.(9)2

-135=9vo-405

9v0=270

v0=30m/s

5. Look at the given picture below. Object K does free fall motion and object B thrown upward at the same time. They collide after 2s. Find the initial velocity of object B. (g=10m/s2) Object A does free fall motion

hA=1/2.10.22=20m

hL=v0.t-1/2.g.t2

hL=v0.2-1/2.10.22

hL=2v0-20

hK+hL=80m

20m+hL=80m

2v0-20=60m

v0=40m/s

Kinematics Exam3 and Problem Solutions

1. As you can see from the given picture, ball is thrown horizontally with an initial velocity. Find the time of motion. (g=10m/s2) Ball does projectile motion in other words it does free fall in vertical and linear motion in horizontal. Time of motion for horizontal and vertical is same. Thus in vertical;

h=1/2g.t2

80=1/2.10.t2

t=4s

2. An object hits the ground as given in the picture below. Find the initial velocity of the object. Velocity of horizontal motion is constant. So;

V0=Vx=Vcos530

Vx=V0=30m/s.0,6

V0=Vx=18m/s

3. An object is thrown with an angle 370 with horizontal. If the initial velocity of the object is 50m/s, find the time of motion, maximum height it can reach, and distance in horizontal. V0x=V0cos530=50.0,8=40m/s

V0y=V0y.sin530=50.0,6=30m/s

a) V-V0y=0-g.t at the maximum height

t=30/10=3s

2.t=time of motion=2.3=6s

b) V0y2=hmax.2.g

hmax=302/2.10=45m

c) X=V0x.ttotal=40.6=240m

4. A balloon having 20 m/s constant velocity is rising from ground to up. When the balloon reaches 160 m height, an object is thrown horizontally with a velocity of 40m/s with respect to balloon. Find the horizontal distance travelled by the object. Object has velocity 40m/s in horizontal, 20m/s in vertical and its height is 160m. We can find time of motion with following formula;

h=V0y.t-1/2.g.t2

-160=20.t-1/2.10.t2

t2=4t-32

(t-8).(t+8)=0

t=8s

X=V0x.t=40.8=320m.

5. Objects A and B are thrown with velocities as shown in the figure below. Find the ratio of horizontal distances taken by objects. Time of flight is directly proportional to vertical component of velocity. Vertical velocity component of A is three times bigger than vertical velocity component of B.

tA/tB=3 tB=tA/3

Horizontal distance traveled by the object is found by the following formula;

XA=VA.tA

XB=VB.tB

Horizontal component of VA is half of VB, so we can write following equation;

VA=VB/2

VB=2.VA

XA=VA.tA

XB=2.VA.tA/3

XA/XB=3/2

Date: 2016-04-22; view: 738

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