Projectile Motion Modelling assumptions Objects thrown in a PROJECTILE MOTION fly in a parabola Modelling assumptions a projectile is a particle (negligible mass) no air resistance no wind not powered

This is no good. Wind and air resistance will effect the motion too much.

Freddie hitting a six

Velocity An object is projected at 40ms-1 at 50o to the x-axis “a measure of speed and direction” An object is projected at 40ms-1 at 50o to the x-axis N What are the horizontal and vertical components of the velocity? 40ms-1 40 sin 50o 50o 40 cos 50o

“Resolving Velocities” E.g a ball thrown with velocity U at degrees to the horizontal U ms-1 U sin   U cos  Horizontal component = U cos  Vertical component = U sin 

Projectile Problems 15 ms-1 “A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the:- time of flight the range the maximum height “ 15 ms-1 30o O x y Maximum Height Range

Equations Horizontal motion Vertical Motion Initial position a u v r

Equations Initial position a -g u v r Horizontal motion Vertical Motion Initial position a -g u v r

Equations Initial position a -g u ux = 15cos30 uy=15sin30 v r Horizontal motion Vertical Motion Initial position a -g u ux = 15cos30 uy=15sin30 v r

Equations Initial position a -g u ux = 15cos30 uy=15sin30 v (=u+at) Horizontal motion Vertical Motion Initial position a -g u ux = 15cos30 uy=15sin30 v (=u+at) vx = vy = r

Equations Initial position a -g u ux = 15cos30 uy=15sin30 v (=u+at) Horizontal motion Vertical Motion Initial position a -g u ux = 15cos30 uy=15sin30 v (=u+at) vx = 15cos30 vy = 15cos30-gt r

Equations Initial position a -g u ux = 15cos30 uy= 15sin30 v (=u+at) Horizontal motion Vertical Motion Initial position a -g u ux = 15cos30 uy= 15sin30 v (=u+at) vx = 15cos30 vy = 15sin30-gt r (=ut+0.5at2) rx= ry =

Equations Initial position a -g u ux = 15cos30 uy= 15sin30 v (=u+at) Horizontal motion Vertical Motion Initial position a -g u ux = 15cos30 uy= 15sin30 v (=u+at) vx = 15cos30 vy = 15sin30-gt r (=ut+0.5at2) rx=15cos30t ry = 15sin30t-.5gt2

Find the time of flight

Find the time of flight When flight is complete y=0 ry = 15sin30t-.5gt2 = 0

Find the time of flight When flight is complete y=0 ry = 15sin30t-.5gt2 = 0 0 = 15 sin 30 t + 1/2 x -9.8 x t2 Either: t = 0 [start] or 7.5 - 4.9t = 0 t = 7.5/4.9 = 1.53 sec. 0 = 7.5t - 4.9t2 0 = t(7.5 - 4.9t)

Find the range A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the range

Find the range A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the range. The range is the horizontal distance travelled before landing.

Find the range A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the range. The range is the horizontal distance travelled before landing. This is the value of x when y=0.

Find the range A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the range. The range is the horizontal distance travelled before landing. This is the value of x when y=0. rx=15cos30t

Find the range rx = 15 cos 30 x 1.53 rx = 12.99 x 1.53 = 19.9 m (1 dp) A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the range. rx = 15 cos 30 x 1.53 rx = 12.99 x 1.53 = 19.9 m (1 dp)

Find the maximum height A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the maximum height

Find the maximum height A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the maximum height Max height occurs when the vertical velocity is zero. This happens at the top of the flight.

Find the maximum height A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the maximum height Max height occurs when the vertical velocity is zero. This happens at the top of the flight. vy = 15sin30-gt 0 = 7.5 - 9.8 t 0 = 15 sin 30 - 9.8 t t = 7.5 / 9.8 = 0.765 sec

Find the maximum height A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the maximum height Max height occurs when the vertical velocity is zero. This happens at the top of the flight. vy = 15sin30-gt 0 = 7.5 - 9.8 t 0 = 15 sin 30 - 9.8 t t = 7.5 / 9.8 = 0.765 sec We now need the value of y at this time

Projectile Problem - so far A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the:- time of flight 1.53 sec the range 19.9 m the maximum height when t = 0.765 Notice the time at the maximum height is exactly half the total time of flight 15 ms-1 30o O x y Maximum Height For ‘symmetrical’ problems like this that will always be the case Range

A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the maximum height t = 0.765 sec ry = 15sin30t-.5gt2

t = 0.765 sec ry = 15sin30t-.5gt2 = 2.87 ~ 2.9 m (1 dp) A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the maximum height t = 0.765 sec ry = 15sin30t-.5gt2 = 2.87 ~ 2.9 m (1 dp)

Projectile Problem - with height elements “A ball is hit with initial velocity 12 ms-1 at an angle of 400 to the horizontal. How long is it more than 2m above the ground ? “ 12 ms-1 40o O x y 2 m

Equations Initial position a -g u ux = 12cos40 uy= 12sin40 v (=u+at) Horizontal motion Vertical Motion Initial position a -g u ux = 12cos40 uy= 12sin40 v (=u+at) vx = 12cos40 vy = 12sin40-gt r (=ut+0.5at2) rx=12cos40t ry = 12sin40t-.5gt2

A ball is hit with initial velocity 12 ms-1 at an angle of 400 to the horizontal. How long is it more than 2m above the ground ? ry = 12sin40t-.5gt2

A ball is hit with initial velocity 12 ms-1 at an angle of 400 to the horizontal. How long is it more than 2m above the ground ? ry = 12sin40t-.5gt2 2 = 12 sin 40 t - 1/2 x 9.8 x t2 2 = 7.713t - 4.9t2 4.9t2 - 7.713t + 2 = 0

A ball is hit with initial velocity 12 ms-1 at an angle of 400 to the horizontal. How long is it more than 2m above the ground ? ry = 12sin40t-.5gt2 2 = 12 sin 40 t - 1/2 x 9.8 x t2 2 = 7.713t - 4.9t2 4.9t2 - 7.713t + 2 = 0 t = 0.33 or 1.25 (2 dp)

t = 0.33 or 1.25 (2 dp) Projectile Problem - with height elements “A ball is hit with initial velocity 12 ms-1 at an angle of 400 to the horizontal. How long is it more than 2m above the ground ? “ t = 0.33 or 1.25 (2 dp) It is above 2m for 1.25 - 0.33 = 0.92 seconds 12 ms-1 40o O x y 2 m