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Aussie NSW Year 12 Projectile Motion Question.

  • Thread starter f3nr15
  • Start date
  • #1
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Homework Statement



A projectile has a time of flight 7.5s and a range 1200m
calculate
(a) its horizontal velocity
(b) Its maximum height
(c) the velocity witch which it is produced

I take ay as -9.8ms-2

Homework Equations



r = uxt
dy = uyt + (ayt2/2)
vy2 = uy2 + 2ah
ux2 + uy2 = u
v = u + at

The Attempt at a Solution



I hope I am the correct

a) r = uxt

ux = r/t
ux = 1200/7.5
ux = 160m

.: Initial Horizontal Velocity is 160m

b)

dy = uyt + (at2/2)

dy = 0 (Total vertical displacement is zero)
0 = 7.5uy - 275.625
7.5uy = 275.625
uy = 36.75

vy2 = uy2 + 2ah

vy = 0
0 = 1350.5625 - 19.6h
19.6h = 1350.5625
h = 68.90625 ...

.: Maximum height is 68.91m-1

c)

ux2 + uy2 = u

(160)2 + (36.75)2 = 26950.5625
u = 164.1662648 ...

.: Initial Velocity is 164.17ms-1

v = u + at

v = (164.1662648) + (-9.8)(7.5)
v = 90.6662648 ...

.: Final Velocity is 90.67ms-1


Ya ?
 

Answers and Replies

  • #2
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,144
486

Homework Statement



A projectile has a time of flight 7.5s and a range 1200m
calculate
(a) its horizontal velocity
(b) Its maximum height
(c) the velocity witch which it is produced

I take ay as -9.8ms-2

Homework Equations



r = uxt
dy = uyt + (ayt2/2)
vy2 = uy2 + 2ah
ux2 + uy2 = u
v = u + at

The Attempt at a Solution



I hope I am the correct

a) r = uxt

ux = r/t
ux = 1200/7.5
ux = 160m

.: Initial Horizontal Velocity is 160m

b)

dy = uyt + (at2/2)

dy = 0 (Total vertical displacement is zero)
0 = 7.5uy - 275.625
7.5uy = 275.625
uy = 36.75

vy2 = uy2 + 2ah

vy = 0
0 = 1350.5625 - 19.6h
19.6h = 1350.5625
h = 68.90625 ...

.: Maximum height is 68.91m-1

c)

ux2 + uy2 = u

(160)2 + (36.75)2 = 26950.5625
u = 164.1662648 ...

.: Initial Velocity is 164.17ms-1

v = u + at

v = (164.1662648) + (-9.8)(7.5)
v = 90.6662648 ...

.: Final Velocity is 90.67ms-1


Ya ?
a, b, and c look correct numerically, but your units generally are not. Then you added a part d which is not correct. You've got to look in the vertical and horizontal direction separately. Since energy is conserved, what might the final velocity be?
 

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