# Difficult Projectile Motion QuestionsWatch

#1
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 .

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?
0
8 years ago
#2
(Original post by Ivo)
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 .

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.

Its the same kind of thing for 2nd one as well
0
8 years ago
#3
2) R = u^2 sin2x / g. For maximum Range R, sin 2x = ?

Equate both R's:

u^2 sin 2x / g = u^2 sin2y/g

sin 2x = sin 2y

(Weird case: imagine if x = 90 deg, y = 0 both will have 0 range.)
0
4 years ago
#4
Part 2 was used in my uni interview

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.
0
4 years ago
#5

The question asks us to figure out the initial velocity (v) and also the time of the flight.
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?
0
3 years ago
#6
The time of flight is given in this question as 0.330 seconds - should make it solvable now.
0
1 year ago
#7
For max range angle=45°Sin2A=1Sin2A=90°SinA= 45°
0
1 year ago
#8
(Original post by Ivo)
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 .

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?

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 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)

For the above equation to be true :-

Either 2A = 2B

- A = B

or

2A = 180- 2B

=> A = 90 - B
0
1 year ago
#9
Q1 is incorrect fool. Did you even put it back into the equation?
The angle is 9.735 degrees
0
1 year ago
#10
(Original post by EMwillcock)
Q1 is incorrect fool. Did you even put it back into the equation?
The angle is 9.735 degrees
Nope, he did it right. I got the same answer (I did it a different way though - his way was a bit faster)
0
1 year ago
#11
i didnt get it
0
1 year ago
#12
got it
0
10 months ago
#13
I got a question. If you say that sin A = 0.6944/2, then sin 30 should equal to sin (60) / 2, isn't it??
0
10 months ago
#14
(Original post by Alpha_Barion)
I got a question. If you say that sin A = 0.6944/2, then sin 30 should equal to sin (60) / 2, isn't it??
You may want to review your understanding on trigonometric ratio.

Note that trigonometric ratios are not linear function.

0
7 months ago
#15
let me explain something in trigonometry sin(2A)=sinAcosA and not 2sinA
0
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