# Help with A Level Question on Projectile Range

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

Hey, here's the question:

An aeroplane carrying out a parcel drop releases a parcel while travelling at a steady speed of 90ms-1 at an altitude of 200m. Calculate:

a) the time between the parcel leaving the plane and it striking the ground.

I worked this out to be 6.39s but correct me if i'm wrong!!

b) the horizontal distance travelled by the parcel in this time.

I'm not sure on this one, the question doesnt say "there's no air resistance" but if the parcel is dropped instantaneously (because the plane is moving, not the parcel?) , surely it only has a vertical 'distance', making the answer 0m.

c) the speed at which the parcel strikes the ground.

Do you simply do s = d/t ? (200/6.39 = 31.3)

or use an equation of motion... i.e

v= u + at

v= 0 + 9.81 x 6.39

v= 62.7ms

Help Help please

An aeroplane carrying out a parcel drop releases a parcel while travelling at a steady speed of 90ms-1 at an altitude of 200m. Calculate:

a) the time between the parcel leaving the plane and it striking the ground.

I worked this out to be 6.39s but correct me if i'm wrong!!

b) the horizontal distance travelled by the parcel in this time.

I'm not sure on this one, the question doesnt say "there's no air resistance" but if the parcel is dropped instantaneously (because the plane is moving, not the parcel?) , surely it only has a vertical 'distance', making the answer 0m.

c) the speed at which the parcel strikes the ground.

Do you simply do s = d/t ? (200/6.39 = 31.3)

or use an equation of motion... i.e

v= u + at

v= 0 + 9.81 x 6.39

v= 62.7ms

Help Help please

0

reply

Report

#2

(Original post by

Hey, here's the question:

An aeroplane carrying out a parcel drop releases a parcel while travelling at a steady speed of 90ms-1 at an altitude of 200m. Calculate:

a) the time between the parcel leaving the plane and it striking the ground.

I worked this out to be 6.39s but correct me if i'm wrong!!

b) the horizontal distance travelled by the parcel in this time.

I'm not sure on this one, the question doesnt say "there's no air resistance" but if the parcel is dropped instantaneously (because the plane is moving, not the parcel?) , surely it only has a vertical 'distance', making the answer 0m.

c) the speed at which the parcel strikes the ground.

Do you simply do s = d/t ? (200/6.39 = 31.3)

or use an equation of motion... i.e

v= u + at

v= 0 + 9.81 x 6.39

v= 62.7ms

Help Help please

**Jess!ca**)Hey, here's the question:

An aeroplane carrying out a parcel drop releases a parcel while travelling at a steady speed of 90ms-1 at an altitude of 200m. Calculate:

a) the time between the parcel leaving the plane and it striking the ground.

I worked this out to be 6.39s but correct me if i'm wrong!!

b) the horizontal distance travelled by the parcel in this time.

I'm not sure on this one, the question doesnt say "there's no air resistance" but if the parcel is dropped instantaneously (because the plane is moving, not the parcel?) , surely it only has a vertical 'distance', making the answer 0m.

c) the speed at which the parcel strikes the ground.

Do you simply do s = d/t ? (200/6.39 = 31.3)

or use an equation of motion... i.e

v= u + at

v= 0 + 9.81 x 6.39

v= 62.7ms

Help Help please

0

reply

Report

#3

for q1 i did -

v^2 = u^2 + 2as

so

90^2 + 2x9.81x200

v = 109.65

then

v = u + at

v-u / 9.81 = t

= 2.00 seconds approx

v^2 = u^2 + 2as

so

90^2 + 2x9.81x200

v = 109.65

then

v = u + at

v-u / 9.81 = t

= 2.00 seconds approx

0

reply

Report

#4

a) Is right

b) The horizontal velocity is 90ms-1 and is constant. Use the time calculated in a) to find horizontal distance

c) Draw a vector diagram of horizontal (90ms-1) and vertical (final vertical velocity) and find resultant vector

b) The horizontal velocity is 90ms-1 and is constant. Use the time calculated in a) to find horizontal distance

c) Draw a vector diagram of horizontal (90ms-1) and vertical (final vertical velocity) and find resultant vector

0

reply

Report

#5

(Original post by

for q1 i did -

v^2 = u^2 + 2as

so

90^2 + 2x9.81x200

v = 109.65

then

v = u + at

v-u / 9.81 = t

= 2.00 seconds approx

**boboyt**)for q1 i did -

v^2 = u^2 + 2as

so

90^2 + 2x9.81x200

v = 109.65

then

v = u + at

v-u / 9.81 = t

= 2.00 seconds approx

0

reply

Report

#8

I just got the answers right

Q:1 ans-

u=0 m/s(initial vertical velocity), s(h)=200m, g=9.81m/s^2

therefore , s=ut+(1/2)at^2.....then solve it the answer is t=6.4s

Q:2 ans-

HORIZONTALLY: u=90m/s, t=6.4s

then use the formula s=vt......the answer is s=576m

Q:3

find v in vertical motion taking u=om/s, s=200m, t=6.4s,v=?m/s

then use the formula s=[(v+u)/2]t to find v.......and the answer will be v=62.5m/s

then Draw a vector diagram of horizontal (90ms-1) and vertical (final vertical velocity) which is 62.5m/s and find resultant vector using Pythagoras Theorem (90^2+62.5^2) whole square root....then the answer will be 109.57...which when rounded will be 110m/s....

JOB DONE !

Q:1 ans-

u=0 m/s(initial vertical velocity), s(h)=200m, g=9.81m/s^2

therefore , s=ut+(1/2)at^2.....then solve it the answer is t=6.4s

Q:2 ans-

HORIZONTALLY: u=90m/s, t=6.4s

then use the formula s=vt......the answer is s=576m

Q:3

find v in vertical motion taking u=om/s, s=200m, t=6.4s,v=?m/s

then use the formula s=[(v+u)/2]t to find v.......and the answer will be v=62.5m/s

then Draw a vector diagram of horizontal (90ms-1) and vertical (final vertical velocity) which is 62.5m/s and find resultant vector using Pythagoras Theorem (90^2+62.5^2) whole square root....then the answer will be 109.57...which when rounded will be 110m/s....

JOB DONE !

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top