Simple Projectile Motion Question (1 Viewer)

[Sonata]

well yea im kinda stuck after encountering my first projectile motion question before starting yr 12 physics lol

A life boat dropped from plane at 140ms-1 at a height of 1102.5m

a)Calculate the initial velocity of the projectile

b)Initial horizontal and vertical velocity

c) its range

d) final horizontal and vertical velocity

if possible explain in detail since im new to this type of thing, btw what does calculate its range supoz to mean?

 

[watatank]

well yea im kinda stuck after encountering my first projectile motion question before starting yr 12 physics lol

A life boat dropped from plane at 140ms-1 at a height of 1102.5m

a)Calculate the initial velocity of the projectile

b)Initial horizontal and vertical velocity

c) its range

d) final horizontal and vertical velocity

if possible explain in detail since im new to this type of thing, btw what does calculate its range supoz to mean?

a) 140ms^-1 in horizontal direction.

b) initial horizontal: 140ms^-1, initial vertical: 0.

c) range is horizontal distance. height is vertical distance. just to clarify. anyways to do this question you need to find time first.

s = ut + 0.5at^2

in vertical direction u = 0, and you know a = 9.8, s = 1102.5 so solve for t (i lack a calculator so do this yourself).

when you have found time, consider s = ut + 0.5at^2 in the x direction. there is no acceleration so s = ut .... you know that u = 140, and you found what t was from the previous q now just solve for s.


d) final horizontal: well in horizontal direction there is no acceleration, so its the same as initial. so 140ms^-1

final vertical: a = ( v - u)/t you have all the info you need, just solve for v.

 

[Sonata]

for
b) after i did this part i found my answers to be the other way around
since it drops from a plane in a straight motion its 90 degrees
so
Vertical = 140Cos90
= 140
Horizontal = 140Sin90
= 0

so what have i done wrong in this part?

c) in the formula s = ut + 0.5at^2
what does the s and t mean?

 

[shredinator]

i don't know hwat you're doing with that trig there.

At the moment the object is released, its velocity is zero in the vertical direction.

But its sitting on a plane that is hurtling along at 140 metres a second, so its horizontal motion is 140ms

 

[watatank]

c) in the formula s = ut + 0.5at^2
what does the s and t mean?

's' is displacement

't' is time

its on your formulas sheet

 

[jlnWind]

for
b) after i did this part i found my answers to be the other way around
since it drops from a plane in a straight motion its 90 degrees
so
Vertical = 140Cos90
= 140
Horizontal = 140Sin90
= 0

so what have i done wrong in this part?

c) in the formula s = ut + 0.5at^2
what does the s and t mean?

Im given year 11 mathematics im sure you would've done work on parabolas?
1st tip to note when you do projectile motion questions
(as in 3U hsc maths and 4U)
is graphing or sketching is always helpful.
2nd that formula as u might notice is a parabola in it self
"s" is the y intercept, "t" is the horizontal axis, so finding the values for t
is like finding the x-intercepts, i.e. one for where the projectile was launched from
and one for where the projectile was launched to (the range).