Projectile problem?

Question

Best answer to whoever has the correct answer and best explanation.

A ball is launched with an initial velocity of 5m/s from a desk with a height of 1.647m above the ground at a 40 degree angle of elevation. How far from the point of launch will it land?

Note, air resistance can be neglected, but if you know a way to igure it in, that would help. If I am missing something needed to answer this, please tell me.

Answer 1

I don't answer a lot of questions, but when I do, I try to give a good, detailed answer.

Air resistance be neglected. There are two parts to this problem: (1) how long will the ball stay in the air, and (2) how far will it go horizontally while it stays aloft?

To answer those questions, we break the problem into vertical and horizontal components. The initial vertical velocity is

vo = 5 sin 40 = 3.21394 m/s (I carry 5 decimals and round off later)

and the (constant) horizontal velocity is

vh = 5 cos 40 = 3.83022 m/s

The general distance formula in the earth's gravitational field is

s = (1/2) gt^2 + vo t + so

where g = -9.8 m/s/s, t is time, vo = 3.21394 is initial velocity, and so = 1.647 m is the initial displacement.

The ball stays in the air until it hits the ground (s=0). So we need to solve the quadratic

(1/2)(-9.8) t^2 + 3.21394 t + 1.647 = 0
-4.9 t^2 + 3.21394 t + 1.647 = 0

Using the quadratic formula,

t = (-3.21394 +/- 6.52768) / (-9.8)

To get a positive root, we must use the negative square root.

t = 0.99404 seconds

The second part of the problem is easy. The constant horizontal velocity is vh = 3.83022 m/s, and you just have to multiply velocity times time. When you do that, you get a horizontal distance of 3.80740 m.

I'd round that off and say the ball landed 3.8 meters from the base of the desk, after a one-second flight. (Answer)


Now, in light of your question, there's one other thing I'd add. You asked how far it landed from . Since it was launched from a desktop, you can use the Pythagorean Theorem:

Total Distance = sqrt(3.8074^2 + 1.647^2) = 4.14836 m, which I'd round off to 4.15 meters if that's what you want. (Answer)

Answer 2

The distance it will land is the time that the ball touches the ground.
The time that the ball will land is the time that the ball reaches negative 1.647m from desk.
The initial vertical velocity is 5xsin40
The initial horizontal velocity is 5xcos40
And you know the acceleration remains -9.8m/s^2 for vertical motion.
Since no force is applied any more to the horizontal, there's no horizontal acceleration.
Vi=initial velocity
Vf=final velocity
d=distance
a=acceleration
Vf^2=Vi^2+2ad
Vf^2=(5xsin40)^2+2x-9.8x-1.647
Vf^2=3.21^2+2x9.8x1.647
Vf^2=42.61
Vf= -6.53m/s
t=(Vf-Vi)/-9.8
t=1second
since it will fall in one second, the horizontal distance it traveled
must be the horizontal speedx time
5xcos40x1=3.8m

Answer 3

first, take the height of the table as the displace in y-component
the you use the equation s=ut-1/2 gt2 where u is initial velocity (must resolve the velocuty given into component), and 2 is gravitational acceleration. put all the values to get the value of t(stands for time).
the you use equation s=ut for x-component where s stands for displacement in x-component. put the values of t from previous equation and the value of u(subscript x) (this also need to resolve)
thes you will get the answer for s.

Answer 4

wat abt the mass of the object.