For Chemistry Tutorials
Introduction
 
Vectors
 
Mechanics
 
Energy Work Power
 
Impulse Momentum
 
Rotational Motion
 
Optics
 
Properties Of Matter
 
Heat Temperature And Thermal Expansion
 
Electrostatics
 
Electric Current
 
Magnetism
 
Waves
 
Exams and Problem Solutions
 
--Vectors Exam and Problem Solutions
 
--Kinematics Exams and Problem Solutions
 
--Kinematics Exam1 and Problem Solutions
 
--Kinematics Exam2 and Problem Solutions
 
--Kinematics Exam3 and Problem Solutions
 
--Kinematics Exam4 and Problem Solutions
 
--Dynamics Exams and Problem Solutions
 
--Work Power Energy Exams and Problem Solutions
 
--Impulse Momentum Exams and Problem Solutions
 
--Rotational Motion Exams and Problem Solutions
 
--Optics Exams and Problem Solutions
 
--Properties of Matter Exams and Problem Solutions
 
--Heat Temperature and Thermal Expansion Exams and Problem Solutions
 
--Electrostatics Exams and Problem Solutions
 
--Electric Current Exams and Problem Solutions
 
--Magnetism Exams and Problem Solutions
 
--Waves Exams and Problem Solutions
 
Old Version
 


Menu

Kinematics Exam3 and Problem Solutions


 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


Author:


Tags:


© Copyright www.PhysicsTutorials.org, Reproduction in electronic and written form is expressly forbidden without written permission of www.PhysicsTutorials.org. Privacy Policy