Some very difficult extension questions on projectile motion, I'll be very impressed if anyone can get these:
1) A golf ball is hit at 60ms-1. At what angle should it leave the club in order to travel 250m horizontally? You will need to use the double angle formula sin2A=2sinAcosA.
2) Prove that, in the absence of air resistance, the maximum range of any projectile is achieved when it is launched at 45o to the horizontal.
3) How can two projectiles launched with the same speed but at different angles have he same range?
Some very difficult extension questions on projectile motion, I'll be very impressed if anyone can get these:
1) A golf ball is hit at 60ms-1. At what angle should it leave the club in order to travel 250m horizontally? You will need to use the double angle formula sin2A=2sinAcosA.
2) Prove that, in the absence of air resistance, the maximum range of any projectile is achieved when it is launched at 45o to the horizontal.
3) How can two projectiles launched with the same speed but at different angles have he same range?
for the 1st one start with , resolving velocities into 60cosx , and 60sinx ..
and consider motion in vertical plane find the time taken
use this time and horizontal distance and speed (60cosx) to find x.
Personally I think the questions are in the wrong order, once you get the distance as a function of angle and velocity for part 2 questions 1/3 become fairly trivial.
The question asks us to figure out the initial velocity (v) and also the time of the flight. I have made two equations: v = velocity, t = total time of flight, 1. v cos(theta) * t = 20 2. 3 = v sin(theta) * t + ((10 t^ 2)/2) Any help?
Some very difficult extension questions on projectile motion, I'll be very impressed if anyone can get these:
1) A golf ball is hit at 60ms-1. At what angle should it leave the club in order to travel 250m horizontally? You will need to use the double angle formula sin2A=2sinAcosA.
2) Prove that, in the absence of air resistance, the maximum range of any projectile is achieved when it is launched at 45 o to the horizontal.
3) How can two projectiles launched with the same speed but at different angles have he same range?
THE FIRST QUESTION'S ANSWER
Range =U^2 sin 2A/ g Take g = 10 ms-2 250 = 60^2 sin 2A/ 10 250*10 = 3600 sin 2A 2500/3600 = sin 2A 0.6944/ 2 = sin A 0.3472 = sin A 0.3472*sin-1 = A A = 20.31
THE SECOND QUESTION'S ANSWER
The maximum height occurs when the vertical speed - Vy = 0 V0y = V0sinθ As a function of time, t: Vy = V0sinθ−gt When Vy = 0: (1):V0sinθ = gt Now horizontal displacement: x = V0xt = V0cosθt
(2):t = xV0cosθ
Substituting for t from (2) in (1): V0sinθ = gx/ V0cosθ =2(V0)^2cosθsinθg
The range is twice this distance so: xmax=2(V0)^2cosθsinθ/g
Substituting the trigonometric identity:2sinθcosθ=sin2θ gives:
xmax=(V0)^2sin2θ/g
Since sin2θ is maximum at 2θ=π/2 then maximum horizontal displacement occurs at θ=π/4 This proves that the maximum range is achieved when it is launched at 45 degree angle
THE THIRD QUESTION
This can happen if the angles at which the objects are launched at are complementary let me explain this with an equation
the range is proportional to sinAcosA = sin 2A
this is shown as
sin(180−A)=sinA
sin2A=sin2(90−A)
So the range of the projectiles is directly proportional to the sine of twice the angle and if everything else was constant the range will be equal when
sin (2A) = sin (2B) ( where A and B are different angles)