2 dimensional motion problem

[demonelite123]

you are sliding down a slide at constant speed of 2m/s. the angle the slide makes with the ground is 30 degrees. At some point while you are sliding down, you shoot a gun and the bullet travels horizontally at the constant speed of 5m/s. at that time, you are also 3m above the ground. How far does the bullet go? (what is the bullets horizontal distance traveled?)

based on how i drew my diagram, i drew a triangle so that i am sliding down to the right. so the triangle has its hypotenuse facing the right side.

so i did 5<1,0> + 2(cos150, sin150> = <3.26, 1>
then i did -3 = t - 4.9t^2

i used quadratic equation, solved for t, and got t=.687. then i plugged t into the equation delta x = 3.26t so the distance the bullet traveled is around 2.24m. i tried using cos30 and sin30 instead and i got 4.63m. which is the correct answer? and how do you know which angle to use?

 

[rl.bhat]

then i did -3 = t - 4.9t^2
This is wrong, because when the bullet is fired its velocity in the downward direction is zero.
When the gun is fired, it was moving down with a uniform velocity. So its horizontal component will add up to the velocity of the bullet.

 

[demonelite123]

then i did -3 = t - 4.9t^2
This is wrong, because when the bullet is fired its velocity in the downward direction is zero.
When the gun is fired, it was moving down with a uniform velocity. So its horizontal component will add up to the velocity of the bullet.

-3 = t - 4.9t^2 is the equation for the y components of the bullet vector and the vector of the person sliding down the slide. since the person is shooting the gun while going down the slide, doesn't the bullet get a vertical velocity and a horizontal velocity and not just horizontal?

 

[latrocinia]

the vector with magnitude two, can be separated in two vectors

2 cos 30' for the y component = 1,73
2 sin 30' for the x component = 1 ((1^2 + 1,73^2) ^ (1/2) = 2)

so the total y component of the bullet is 1,73+ 5 = 6.73m/s

it takes 3 seconds for the x-component to equal 3 so it has 3 seconds to travel

3*6.73 = 18 + 2,19 = 20,19 mor when the bullet just has a velocity of 5m/s it travels only 15m. it's not really clear if the 5m/s velocity is it relative velocity or total.

 

[demonelite123]

the vector with magnitude two, can be separated in two vectors

2 cos 30' for the y component = 1,73
2 sin 30' for the x component = 1 ((1^2 + 1,73^2) ^ (1/2) = 2)

so the total y component of the bullet is 1,73+ 5 = 6.73m/s

it takes 3 seconds for the x-component to equal 3 so it has 3 seconds to travel

3*6.73 = 18 + 2,19 = 20,19 m


or when the bullet just has a velocity of 5m/s it travels only 15m. it's not really clear if the 5m/s velocity is it relative velocity or total.

how come you used cos and sin 30 and not 150? does it matter which angle you use?