Projectiles colliding on a horizontal plane

Woolyabyss
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Homework Statement


A bullet is fired from a gun fixed at a point O with speed v m/s at an angle Θ to the horizontal, At the instant of firing , a moving target is 10m vertically above O and traveling with constant speed 42√ 2 m/s at a constant angle 45 degrees to the horizontal.The bullet and target move in the same plane.

(i) if v = 70 m/s, show that if the bullet is to hit the target then tanΘ = 4/3

(ii) Find at what time after firing does the bullet strike the target and calculate the horizontal distance of the bullet from O


Homework Equations



s=ut + (1/2)gt^2


The Attempt at a Solution



If they are to collide then there horizontal velocities must be equal.

70cosΘ = 42√2(cos45) .... cosΘ = 42/70 =3/5

using Pythagoras' sinΘ = 4/5 and tanΘ = 4/3


(ii) There height must be the same when they collide h1 = h2

42√ 2(sin45)t - (1/2)gt^2 + 10 = 70(4/5)t - (1/2)gt^2

(1/2)gt^2 cancel

42t + 10 = 56t ...... t = 10/14 seconds

horizontal distance of the bullet from O 70(3/5)(10/14) = 30m

My book says the answers for (ii) are 10/7 seconds and 60m.

Can anybody spot if I went wrong at any point? Any help would be appreciated.
 
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Woolyabyss said:
(ii) There height must be the same when they collide h1 = h2

42√ 2(sin45)t - (1/2)gt^2 + 10 = 70(4/5)t - (1/2)gt^2

The target is not in free fall.
 
Thanks I removed that -(1/2)gt^2 and got the right answer.
 

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