# Help with A Level Question on Projectile Range

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#1
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
10 years ago
#2
(Original post by 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 Im pretty sure the parcel moves in a parabolic shape still, so use the equations of motion (s=ut) hope that helps! 0
10 years ago
#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
0
10 years ago
#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
0
10 years ago
#5
(Original post by 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
You've used horizontal velocity which does not change.
0
10 years ago
#6
oops didnt realise u = 0 so your right yh
0
#7
Sweet. Thank youu and Q3?
0
7 years ago
#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 !
0
2 years ago
#9
(a) how can we use s=ut 1/2 at^2 and we don't have t??
0
4 weeks ago
#10
i found my answer to be 6.39s as time
0
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