Projectile motion of a stuntman

[stupif]

1. a stuntman performs a stunt of speeding off the cliff of a high rise building with the aim reaching another building with a distance away as the distance between two building is 62m and the differences height between two building is 48m. find the minimum velocity that the car must have as it reached the edge of the roof in order for the stunt to be successful.



2. first, i find the distance of car from the building to another building = 78.41m
48^2 + 62^2 = (distance traveled by car)^2
distance traveled by car= 78.41m
after that what equation should i use?? and can i assume final velocity = 0m/s?
help me...thank you

The Attempt at a Solution

 

[tiny-tim]

2. first, i find the distance of car from the building to another building = 78.41m
48^2 + 62^2 = (distance traveled by car)^2
distance traveled by car= 78.41m

sorry, but this is pointless: the car won't move in a straight line, will it?

you need to use the standard https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations, once in the x direction with acceleration zero, and once in the y direction with acceleration -g …

what do you get? ​

 

[stupif]

but the problem is what is the final velocity? zero or other value??
i know the car falling dowan at 37.75degree from horizontal. i use this formula, tan angle = 48m/62m, hence equal to 37.75degree.

 

[tiny-tim]

i know the car falling dowan at 37.75degree from horizontal. i use this formula, tan angle = 48m/62m, hence equal to 37.75degree.

sorry, but this is pointless: the car won't move in a straight line, will it?

do you know the standard https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations?​

 

[stupif]

you mean should be parabolic path??
ya... i know...
v^2=u^2+2as
v=u +at
s=ut+at^2

 

[tiny-tim]

you mean should be parabolic path??
ya... i know...
v^2=u^2+2as
v=u +at
s=ut+at^2

(try using the X² tag just above the Reply box )

yes

(but the last one should be s=ut+ (1/2)at²)

ok, now do that horizontally, with a = 0, and vertically, with a = -g …

what do you get? ​

 

[stupif]

sorry...i know the equation...too panic...
you mean find the v right now?but i don't know the t.
and i think a= +g, because toward the ground.
the correct answer of this question is 19.8m/s.

 

[tiny-tim]

you mean find the v right now?but i don't know the t.

you will have two equations (one horizontal and one vertical), and two unknowns (v and t) …

that should be enough to solve it!

write out the two equations ​

 

[stupif]

i tried b4
s= vt, s=ut+1/2at^2
t is same
v is same,
so,
s=u(s/v) + 1/2a(s/v)^2
48m= 62m+ 1/2(9.8) (3844/v^2)
v=36.68m/s, the correct answer should be19.8m/s

 

[tiny-tim]

s=u(s/v) + 1/2a(s/v)^2

yes, that's correct

48m= 62m+ 1/2(9.8) (3844/v^2)

no, this is the y equation, so the 48 is correct, but the next term should be u_yt, not x

(u_y is the initial component of velocity in the y direction)

try again ​

 

[stupif]

sorry， i do not understand...is the horizontal initial speed is not same as vertical initial speed?

 

[tiny-tim]

no no nooo …

the car is initially going at speed v along a flat roof, so the horizontal component is v, and the vertical component is zero

generally, if the velocity is speed v at an angle θ to the horizontal,

then the horizontal component of velocity is vcosθ, and the vertical component is vsinθ ​

have you not done components of vectors?

 

[stupif]

having dizzy...
still can't figure it out...

 

[tiny-tim]

u_y = 0 (it has no initial vertical speed)

so u_yt = 0

so … ?​

 

[stupif]

ya...s =vt...horizontal moving in the constant velocity. this one i got it.

 

[tiny-tim]

ya...s =vt...horizontal moving in the constant velocity. this one i got it.

yes, s = u_xt, which in this case is x = vt, from which you've already got your t = x/v

but that's the horizontal component equation …

i'm talking about the vertical component equation, y = u_yt + 1/2 gt²

 

[stupif]

then you mean 48= 0 +1/2(9.8)(62/v cos 37.75)^2
if like this, i also can't get the answer

 

[tiny-tim]

then you mean 48= 0 +1/2(9.8)(62/v cos 37.75)^2
if like this, i also can't get the answer

you're losing the plot …

the plot is, first you find t from the x equation (you got that right, t = x/u_x = x/v),

then you find v from the y equation (using u_y = 0)

the 37.75° is irrelevant ​

 

[stupif]

but i don't know what is v on horizontal line...s=vt, v=?

 

[tiny-tim]

i don't understand …

you've already done this correctly …

the car is initially moving horizontally, so u_x = vcos0° = v, and x = u_xt

 

[stupif]

hah?
then t = 62s?
then s= 1/2at^2, where is the v?

 

[tiny-tim]

v = 62/t (from the x equation), and now you use the t you've just found from your y equation

 

[stupif]

ya...thank you...i got it...
i feel quite disappointed to me, because can't solve the question.
any guidances while doing projectile motion's question?

 

[tiny-tim]

any guidances while doing projectile motion's question?

yup …

always write the formulas out carefully beforehand (ie without any numbers),

and use the u_x and u_y notation instead of just u,

and x or y instead of s …

then there's less chance of getting confused half-way through, and the equations are more likely to mean something to you

 

[stupif]

actually i quite confuse with this, sin and cos?? when i should use sin angle and cos angle??

 

[tiny-tim]

it is always cos

if θ is the angle between the vector V and the direction you want, the the component of V in that direction is always Vcosθ

(the only time you use Vsinθ is when the angle called θ is the wrong angle …

eg if you're told that V is at an angle θ to the horizontal, and you want the vertical component, then the angle between V and the vertical is 90° - θ, so the component is Vcos(90° - θ), which is Vsinθ )