Projectile Motion - Time in Air

[air-in]

Homework Statement

What angles should an arrow be launched if it leaves the bow @ speed of 60 m/s and is trying to hit a target that is 150 m away? Assume target is same height as bow.
Suppose the arrow is fired at an angle of 30 degrees & misses the target & hits the ground 1.8 m below from where it started. How much time was it in the air?

Homework Equations

1. t = 2vsinθ/g
2. t = √2y/g
3. Δy = (vsinθ)^2/2g

The Attempt at a Solution

I have attempted this the bold section of this problem a few different ways, but I am not convinced that any of them are correct. Could someone please tell me if any of them are correct, and if not, explain to me how to do it.

Attempt 1: t = 2(60)sin30/9.8
t = 6.1s

Attempt 2: t = √2(1.8)/9.8
t = .61s

Attempt 3: 1.8 = (vsin30)^2/2(9.8)
v = 8.4 m/s
t = 2(8.4)sin30/9.8
t = .86s

 

[_N3WTON_]

Have you remembered to take into account the fact that the arrow lands 1.8m below it's original launch height?

 

[air-in]

Have you remembered to take into account the fact that the arrow lands 1.8m below it's original launch height?

Wouldn't that be Δy, which I tried using in attempts 2 and 3?

 

[_N3WTON_]

Now that you have edited your solution I believe that you are correct..

 

[HalfLostAndSearching]

Attempt 4: 1.8 = (60sinθ)^2/2(9.8)
θ = 23.7 degrees
t = 2(60)sin23.7/9.8
t = 4.2s
To hit a target 150 m away with a speed of 60 m/s, the arrow should be launched at an angle of approximately 45 degrees. This is because at this angle, the horizontal distance traveled by the arrow will be equal to the distance to the target. However, this does not take into account air resistance and other factors that may affect the trajectory of the arrow.

For the second part of the problem, the arrow was fired at an angle of 30 degrees and missed the target, hitting the ground 1.8 m below its starting point. To find the time in air, we can use the equation t = √2y/g, where y is the vertical displacement and g is the acceleration due to gravity. Plugging in the given values, we get t = √2(1.8)/9.8 = 0.61 seconds.

Your first attempt is incorrect because it does not take into account the vertical displacement of the arrow.

Your second attempt is correct, as it uses the correct equation and takes into account the vertical displacement of the arrow.

Your third attempt is incorrect because it assumes that the arrow is launched at an initial velocity of 60 m/s, which is not the case. The arrow is launched at a speed of 60 m/s, but at an angle of 30 degrees, the initial velocity in the horizontal direction is only 30 m/s.

Your fourth attempt is incorrect because it uses the wrong angle. The arrow was fired at an angle of 30 degrees, not 23.7 degrees.