1. OK so if you imagine a projectile is fired with an inital velocity (u) of 400 m/s at an angle(adsbygoogle = window.adsbygoogle || []).push({}); theta. What two values ofthetawill cause the projectile to hit a point b that is 5 km horizontally away from the point of launch and 1.5 km vertically?

u (magnitude)= 400 m/s

deltaH = 1500 m (1.5 km)

deltaS = 5000 m (5 km)

2. constant acceleration equations, trigonometric identities.

3. What i tried to do was:

u_{x}=400cos(theta)

u_{y}=400sin(theta)

using s=ut+1/2at^{2}

1500 = 400t.Sin(theta)-4.905t^{2}(1)

5000 = 400t.cos(theta)

t = 12.5/cos(theta) (2)

substituting (2) into (1) gives,

1500 = 5000Sin(theta)/Cos(theta) - 766.4/Cos^{2}(theta)

multiplying by Cos^{2}(theta) gives

1500cos^{2}(theta) = 5000sin(theta)cos(theta) - 766.4

= 2500Sin(2theta) - 766.4

= 5000Cos^{2}(theta) - 2500 - 766.4

3500cos^{2}(theta) = 3276

cos^{2}(theta) = 0.9361

cos(theta) = 0.9675

theta = 14.6°

which is the wrong answer. Obviously i'm missing something here, but i don't know what. Hence why i'm posing the question on here. Thanks EH.

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# Projectile motion problem involving finding launch angle

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