Pradeep Kshetrapal is teaching live on Unacademy Plus
PHYSICS Pradeep Kshetrapal Class XI
Chapter 4 Motion in a plain Projectile Motion Lesson No.Phy-XI-4-7
The next step of finding relation between quantities in two dimensional motion is to use equations of kinematics and applly them separately for horizontal and vertical direction.
In horizontal way the acceleration due to gravity s zero So, distance covered is s- V x t In the horizontal, during the whole flight velocity is same as initial velocity. So, horizontal velocity remains constant through out the flight.
In vertical direction the velocity is accelerated velocity. Acceleration ag. The equations in vertical component are VtUSin-gt
Different parameters of projectile motion. TIME OF FLIGHT Trajector)y Hmax ucos 0 Range Flight of projectile is continued only till object is above the ground therefore the time of flight depends on vertical motion.
But, in horizontal direction we don't know the displacement so, we cant use the formula for displacement-Velocity x time. Remember, Time of flight is always decided by vertical motion. In vertical motion the object reach a certain height and again hits the ground. So the total displacement is zero.
By the equation s- ut + at2 2 (In vertical direction.) Initial velocity- usin 0 Acceleration-2. Final vertical Displacement- 0 Time =T.
2 2 Divide both sides with T 1 2u sin Time of flight T-usine
Maximum height achieved, Hmax Condition for maximum height is that at that point vertical velocity ecomes zero initial vertical velocity-u sin acceleration- -g final vv-0 S= Hmax = ? max
Height is maximum when Sin is maximum or -90. Then Hmax-u?sin290-42 Then Hmax 2g2g
Range is maximum when Sin20-1 (Maximum). So, 20-90. or =45. Hence at 45 degree projection the range is maximum This maximum range is 2 max
For any given velocity, maximum range is double the maximum height Maximum range ( at 45)- Rmax- ^ At this angle 45 the maximum 2 height is Hmax Sin290 U2 2g