HELP PLEASE..

on level ground a shell is fired with an initial velocity of 50.0 m/s at 60.0 degrees above the horizontal and feels no appreciable air resistance. a.) Find the horizontal and vertical component's of the shells initial velocity. b.)How long it take the shell to reach its highest point?. c.) Find its maximum height above the ground.d.) How far from its firing point does the shell land?. e.) At its highest point,find the horizontal and vertical components of its acceleration and velocity.

Question

HELP PLEASE..

on level ground a shell is fired with an initial velocity of 50.0 m/s at 60.0 degrees above the horizontal and feels no appreciable air resistance. a.) Find the horizontal and vertical component's of the shells initial velocity. b.)How long it take the shell to reach its highest point?. c.) Find its maximum height above the ground.d.) How far from its firing point does the shell land?. e.) At its highest point,find the horizontal and vertical components of its acceleration and velocity.

anonymous · Answer

@ganeshie8

tacotime · Answer

Horizontal velocity = initial v * cos (angle)
Vertical velocity = initial v * sin (angle)

anonymous · Answer

horizontal= -47.62
vertical= -15.24

anonymous · Answer

How long it take the shell to reach its highest point?

tacotime · Answer

I know that at the highest point the horizontal component is still 15.24 because there is no horizontal acceleration.
Other than that I'm stuck, sorry.

anonymous · Answer

do you know anyone that can help? ^^

tacotime · Answer

@Michele_Laino seems to be the only knowledgeable person tonight

michele_laino · Answer

the horizontal component of velocity is:
$${v_x} = {v_0}\cos \theta  = 50 \cdot \cos 60 = ...?$$

michele_laino · Answer

whereas the vertical component of velocity is:
$${v_y} = {v_0}\sin \theta  = 50 \cdot \sin 60 = ...?$$

anonymous · Answer

horizontal= -47.62
vertical= -15.24

anonymous · Answer

How long it take the shell to reach its highest point?

zephyr141 · Answer

are you sure you put those numbers into your calculator correctly? i mean the shell is fired 60 degrees above the horizontal. i'm getting positive numbers for my y and x directions.

zephyr141 · Answer

first of all you should always start by drawing a FBD|dw:1444195069945:dw|

anonymous · Answer

yup i checked it..

zephyr141 · Answer

so at the highest point you have no velocity in the y direction. so by using the y component you found earlier solve for when v final is 0. you have v initial. look at the kinematic equations and find the one you can work with right now.$$d=v_0t+\frac{1}{2}at^2$$$$v^2=v_0^2+2ad$$$$v=v_0+at$$$$d=\frac{v_0+v}{2}t$$ i suggest finding time first then finding the distance.

anonymous · Answer

we can use the first formula you gave?

zephyr141 · Answer

not yet. look at it. there's 5 variables. we have three. initial velocity, final velocity, and acceleration. we need the distance (which is what we're trying to find) and time (which can be found using another equation). look at the equations and see which one we can find time.

anonymous · Answer

v2=v20+2ad

zephyr141 · Answer

also you're getting those numbers because your calculator is in Radians mode, you'll want to be in degree mode to get the answers of 25 for velocity in the x-direction and 43.3 for the y direction.

anonymous · Answer

i switch it in radians mode and i still got the same value

anonymous · Answer

hmm, ok i get it ^^

zephyr141 · Answer

nah you're in radians mode right now. that's why you're getting negative values. switch to degree mode. i mean i just checked with my CASIO fx-9750GII and confirmed it. you're in radians mode. you need to be in degree mode.