Question

Example 4!
A particle is projected from a point which is 2m above ground
level with a velocity of 40 m/s at an angle of 45° to the horizontal.
Find its horizontal distance from the point of projection when it
hits the ground.​

Answer 1

Answer:

Horizontal Velocity= 40 x Cos(45)= 28.28 m/s

Vertical Velocity= 40 x Sin (45) = 28.28 m/s

v^2= u^2 + 2as

0 = 28.28^2 - 2 x 9.81s

19.62s= 800

s= 40.77 m

s= ((v+u) x t)/2

81.55= 28.28t

t= 2.88 s

Total Distance to fall= 40.77+2 = 42.77m

s = ut+ 0.5at^2

42.77 = 0 +0.5 x 9.81 x t^2

85.55 = 9.81 t^2

t= 2.95 s

Total Time= 2.95 + 2.88= 5.64 s

Total Distance= 5.64 x 28.28= 165.06 m