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Projectile Motion

Published in: AIEEE | IIT JEE Mains | Physics
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  • Harsh K

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    • Qualification: M.Sc
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All The Relevant Concepts Of Projectile Motion ( Kinematics).

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    What do we mean by projectile motion? In kinematics we deal with the motion of a point particle. Projectile motion is nothing but the motion of a point particle under the influence of a unidirectional force like gravitational force close to earth's surface. The two tools to study projectile motion are the i) Equation of motion. ii) Projection of velocity vector.
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    v S 2 u ± a*t u*t ± a*t2/2 — u2 ± 2a*S Equation of motion u - Initial velocity v — Final velocity S — Displacement Acceleration a t - Time Keep in mind the velocity, displacement and acceleration are all vector quantity and their direction will determine sign in the above equations.
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    y-axis Vector projection We see in the figure a vector with magnitude I vl and direction along angle 0 from x-axis. The component of velocity can be resolved along x and y-axis as v x and v respectively. v - v*cos0 v - v*sin0 x-asls
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    Breaking down a Projectile Acceleration due to gravity g— 9.8ms- h d Using the three equations of motion and the vector projection we can treat projectile motion as two separate linear motion. The velocity along x-axis will remain constant as it is not subjected to any external force. The velocity along the y-axis will be affected by gravity thus the particle will go up and then down again. This is the velocity which determines the flight time of the projectile.
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    Time of flight We find the total time of light by using the first equation of motion v = u ± a*t Here, u as the force is opposite to the direction of velocity we use a negative sign. Now since velocity of the particle at maximum height will be O we can find the time taken for it to reach this height by substituting v as 0.
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    Cont, , This gives us v*sin0/g This is only half the total time of flight as the particle comes back to the ground and since the force on it is same, the time will be equal to what it took to reach the height h. Thus, t = 2*v*sin0/g This time will be very useful in our calculations further to calculate the horizontal distance travelled.
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    Maximum height To calculate maximum height we can use methods. Since we know the time taken by the particle to reach the height h, we can simply substitute the value t' in the second equation of motion ± at2/2 Therefore, h=vsin0*vsin0/g - g*v2sin20/2g2 h=v2sin20/2g We can use the third equation of motion v2=u2 ± 2*a*S Substituting the final and initial velocity along y-axis, we get, 0=v2sin20 — 2*g*h h=v2sin20/2g
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    Horizontal distance To calculate the maximum horizontal distance travelled by the projectile, we multiply the velocity along x-axis and the total time of flight. X — v*cos0*t X=v*cos0*2*v*sin0/g X = v2sin20/g
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    h Projectile from a height In this case the particle is thrown from a height h. Thus the time of flight increases than in normal case. However despite the complexities all the equation of motion holds well when used with appropriate Sign. e.g. If h is given and time of flight (t) is given we can find v by using the second equation of motion h = v *sin0*t — g*t2/2 Or if the initial and final velocities are given we can use third law to find the distance and so on.
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    Trajectory of Motion To find the trajectory of a projectile is very useful in certain problems where we have to find if the projectile will hit a target or where will it hit on a wall given the distance of the wall from the starting point etc. To find the trajectory of any motion we need to eliminate t from the equation. We use the second equation of motion Y - v t - gt2/2 x = vxt t=x/v Substituting, y x(v / v) - gx2/2v 2
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    Cont., y = x*(vsin0/vcos0) - g*x2*/2vcos02 y = xtan0 - gx2/2v2cos20 Thus, we see that the trajectory of a projectile is a parabola.
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    END I hope it the presentation helped you clarify some concept of the projectile motion. I will try to bring more content out in the future. Thanks for viewing For any doubt contact me - [email protected]

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