A)Consider the setup of a gun aimed at a target (such as a monkey) as shown in t

Question

A)Consider the setup of a gun aimed at a target

(such as a monkey) as shown in the figure.

The target is to be dropped from the point A

at t = 0, the same moment as the gun is fired.

The bullet hits the target at a point P, which

is at the same horizontal level as the gun.

Let the initial speed of the bullet be v0,

let the angle between the vector ~v0 and the

horizontal (x-) direction be , and OP = L

and AP = h. The gravitational acceleration

is g. Denote the time taken to hit the target

by T.

The acceleration of gravity is 9.8 m/s2 .

B)Find the initial speed v0 (magnitude of the

vector ~v0) which allows the projectile to meet

the target at location P. (Hint: T defined in

part 1 is also the time taken for the bullet

to travel, following the projectile trajectory,

from O to P). Let the distance OP be L =

1.35 m, the angle = 42.4, and the time

T = 0.501573 s. (Given g = 9.8 m/s2).

= 3.64 m/s

C)Now the same setup is to take place at some

planet where the gravitational acceleration is

g' =g/4. Keep v0, and h to be the same as

before. Find the new height; i.e., the y-coordinate

of the new point of collision. (Hint: you

should convince yourself that for this new

case, the time taken for the bullet to travel

from O to the new point of collision P should

still be T).

Please just explain part C!

Explanation / Answer

Of course the equations of physics holds whatever numerical values their variables takes. It means for the planet where we have gravitational acceleration g' the time to target T' is

T' = sqrt(2h/g') = sqrt(8h/g) = 2*sqrt(2h/g) =2T

Since on the vertical there is free fall we have on earth

h =g*T^2/2 (=g*2h/g/2 =h)

and on the planet

h' =g'*T'^2/2 =(g/4)*4T^2/2 =gT^2/2 = h

Therefore the correct answer is 6.

y=h

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