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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.