An object is throwed with an angle of 30º...?

Question

An object is throwed with an angle of 30º and an initlial velocity of 40m/s, if the initial y is 0 in what is the time it is going to take it to y to be 0 again and what is x going to be in that moment?

Give me the formulas plz

Answer 1

Vx = V cos 30 ; Vy = V sin 30 ( V here is the initial velocity , Vx,Vy are its two components on vertical and horizon )
Vy1 = Vy - g*t
y = -1/2 * g*t^2 + Vy*t

Vx1 = Vx ; x = Vx*t

y = 0 =<=> 2 Vy*t - g*t^2 = 0 <=> t = 2Vy/g( since t<>0 )
t = 2 * V sin 30 / g = 4s ( we presume g =10 m/s^2 )
x = Vx * t = V cos 30 * t = 138.56 m

Answer 2

first resolve the initial velo U in to vertical and horizontal components
U(v) = 40 sine30
U(h) = 40 cos 30

Vertical motion V = U -gt
0 = 40cos30 -9.8 t

t = 40cos30 /9.8 timevertical velo will be o m/s

Vertical distance travelled in time t above

S(v) = Ut - g t**2 Please use your calculator
This is also your y value

Downward travel here U =0 S you know from above

S = g t**2 this will give you the time back to ground

Add up two t values for upward and downward travels

Use the total time with horizontal velo. to calculate the distance in horizantal travel. That is also your x Value.

You can use two values of times to find horizontal distance during up motion, and one during down motion.

Hope this helps you

Answer 3

I suggest you copy the problem exactly as it is written in your book. Deciphering what you are trying to say is in itself a problem. You may have to do something about your English to enable you to learn Physics well. Good luck.

Answer 4

the time after which y will be zero is 4s.and value of x at that moment will be 136m.these can be calculated using formula for projectile.