Aussie NSW Year 12 Projectile Motion Question.

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SUMMARY

The projectile motion problem involves calculating the horizontal velocity, maximum height, and initial velocity of a projectile with a time of flight of 7.5 seconds and a range of 1200 meters. The calculations yield an initial horizontal velocity of 160 m/s, a maximum height of 68.91 meters, and an initial velocity of approximately 164.17 m/s. The final velocity, accounting for gravitational acceleration, is approximately 90.67 m/s. The discussion emphasizes the importance of maintaining consistent units throughout the calculations.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Knowledge of projectile motion concepts
  • Familiarity with gravitational acceleration (ay = -9.8 m/s²)
  • Ability to perform basic algebraic manipulations
NEXT STEPS
  • Study the derivation of kinematic equations for projectile motion
  • Learn about energy conservation principles in projectile motion
  • Explore the effects of air resistance on projectile trajectories
  • Investigate advanced projectile motion problems involving angles and initial velocities
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f3nr15
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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 ?
 
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f3nr15 said:

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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