(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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 a_{y}as -9.8ms^{-2}

2. Relevant equations

r = u_{x}t

d_{y}= u_{y}t + (a_{y}t^{2}/2)

v_{y}^{2}= u_{y}^{2}+ 2ah

u_{x}^{2}+ u_{y}^{2}= u

v = u + at

3. The attempt at a solution

I hope I am the correct

a) r = u_{x}t

u_{x}= r/t

u_{x}= 1200/7.5

u_{x}= 160m

.: Initial Horizontal Velocity is 160m

b)

d_{y}= u_{y}t + (at^{2}/2)

d_{y}= 0 (Total vertical displacement is zero)

0 = 7.5u_{y}- 275.625

7.5u_{y}= 275.625

u_{y}= 36.75

v_{y}^{2}= u_{y}^{2}+ 2ah

v_{y}= 0

0 = 1350.5625 - 19.6h

19.6h = 1350.5625

h = 68.90625 ...

.: Maximum height is 68.91m^{-1}

c)

u_{x}^{2}+ u_{y}^{2}= 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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# Homework Help: Aussie NSW Year 12 Projectile Motion Question.

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