# Difficult Projectile Motion Questions

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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

2) Prove that, in the absence of air resistance, the maximum range of any projectile is achieved when it is launched at 45

3) How can two projectiles launched with the same speed but at different angles have he same range?

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?

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#2

(Original post by

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

2) Prove that, in the absence of air resistance, the maximum range of any projectile is achieved when it is launched at 45

3) How can two projectiles launched with the same speed but at different angles have he same range?

**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?

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

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#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.)

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.)

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#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.

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.

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#5

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?

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#6

The time of flight is given in this question as 0.330 seconds - should make it solvable now.

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#8

**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 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)

For the above equation to be true :-

Either 2A = 2B

- A = B

or

2A = 180- 2B

=> A = 90 - B

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#9

Q1 is incorrect fool. Did you even put it back into the equation?

The angle is 9.735 degrees

The angle is 9.735 degrees

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#10

(Original post by

Q1 is incorrect fool. Did you even put it back into the equation?

The angle is 9.735 degrees

**EMwillcock**)Q1 is incorrect fool. Did you even put it back into the equation?

The angle is 9.735 degrees

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#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??

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#14

(Original post by

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??

**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??

Note that trigonometric ratios are not linear function.

This thread is more than 5 years, so you would be better that you start a new thread to ask your query.

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