# Projectile motion of a stuntman

• stupif
In summary, the stuntman must reach a minimum velocity of 36.68m/s at the edge of the roof in order for the stunt to be successful. To find this velocity, the standard equations of motion were used in both the horizontal and vertical directions, with the acceleration being -9.8m/s^2 due to gravity. The initial vertical velocity of the car was assumed to be 0m/s, and the angle of descent was irrelevant to the calculations. It was important to correctly write out the formulas and use the notation of ux and uy for the initial horizontal and vertical velocities, respectively. The use of sin and cos depends on the given angle and the desired component of velocity.
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

stupif said:
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?

Last edited by a moderator:
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.

stupif said:
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?

Last edited by a moderator:
you mean should be parabolic path??
ya... i know...
v^2=u^2+2as
v=u +at
s=ut+at^2

stupif said:
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?

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.

stupif said:
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

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

stupif said:
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 uyt, not x

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

try again

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

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?

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

uy = 0 (it has no initial vertical speed)

so uyt = 0

so … ?​

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

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

yes, s = uxt, 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 = uyt + 1/2 gt2

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

stupif said:
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/ux = x/v),

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

the 37.75° is irrelevant

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

i don't understand

you've already done this correctly …

the car is initially moving horizontally, so ux = vcos0° = v, and x = uxt

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

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

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?

stupif said:
any guidances while doing projectile motion's question?

yup …

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

and use the ux and uy 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

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

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 )

## 1. What is projectile motion?

Projectile motion is the motion of an object in a curved path under the influence of gravity. It occurs when an object is launched into the air and then falls back to the ground, following a parabolic trajectory.

## 2. How does projectile motion apply to a stuntman?

In stunt work, projectile motion is often used to calculate the trajectory of a stuntman's jump or fall. By understanding the principles of projectile motion, stunt coordinators can plan and execute stunts safely and accurately.

## 3. What factors affect the trajectory of a stuntman's jump?

The trajectory of a stuntman's jump is affected by the initial velocity, angle of launch, and air resistance. Other factors such as wind speed and direction can also have an impact on the trajectory.

## 4. How can you calculate the maximum height and distance of a stuntman's jump?

The maximum height and distance of a stuntman's jump can be calculated using mathematical equations based on the initial velocity, angle of launch, and acceleration due to gravity. These calculations can also be simulated using computer software.

## 5. What safety precautions should be taken when considering projectile motion in stunt work?

Stunt coordinators should always consider the potential risks and hazards associated with projectile motion in stunt work. They should conduct thorough risk assessments, provide proper training and protective gear for stunt performers, and have emergency procedures in place in case of accidents.

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