Question

hey guys, i need some help with projectile motion.
i got 2/3 but i need help for the last one

Answer 1

A ball is thrown horizontally at 19 m/s from a clifftop 55 m above the sea level. Find:

a) the time to land; 3.350296971
b) the angle at which it hits the water; 58.77
c) the speed at which it hits the water??

Answer 2

Use conservation of energy, it is the easies approach.

We know from the assumptions associated with projectile motion that \(v_x = {\rm constant}\). We can find the velocity gained in the y-direction from conservation of energy, as\[{1 \over 2} m v_y^2 = mgh \rightarrow {1 \over 2} v_y^2 = gh \rightarrow v_y^2 = 2gh\]
Then the magnitude of both velocity vectors can be found as\[v = \sqrt{v_x^2 + v_y^2}\]

Answer 3

ya but i had found 11.52 for the y and 19 for the x since it is the same.
when you do that you will get around 22.2 and that answer was wrong when i submitted it :S

Answer 4

Double check you \(v_y\). I'm getting \(v \approx 38\) from the magnitude.

Answer 5

but how? cuz i had used those numbers to get my angle

Answer 6

It's simple math. \[v_y = \sqrt{2 \cdot 9.81 \cdot 55} \approx 33\]

Therefore, \[v = \sqrt{19^2 + (2 \cdot 9.81 \cdot 55)} \approx 38\]

The angle is found as\[\tan \theta = {v_y \over v_x}\]

Answer 7

so the angle is 60?

Answer 8

Yes.

Answer 9

odd how it accepted my answer of 58.77 lol
thanks a lot though :)
care to help me with a similar one?

Answer 10

It probably is programmed to allow for a certain amount of % error. 58.77 is less than 3% different than 60.

Answer 11

Did it take 38 as the speed?

Answer 12

yes :)

Answer 13

Sure. Post it up and I'll take a look at it.

Answer 14

Great!

Answer 15

is there another way of solving for it without energy? like using kinematics?

Answer 16

OMG I REALISED MY STUPID MISTAKE. you can use kinematics to solve for vf but that vf will b for the y so then you have to do it for the total!

Answer 17

so stupid :P sry its because i wrote everywhere LOL quiet a big mess :P

Answer 18

Absolutely. \[v_y^2 = 2ad\], which is derived from constant acceleration kinematics. Notice, its the same equation we got when we used energy.

Answer 19

Indeed. The gravitational acceleration only acts in the vertical direction.

Answer 20

Take the time to work through problems in a very organized manner. It's a pain, I know, and uses a ton of paper, but it will help you not silly mistakes like that. :)

Answer 21

ya true :P
i was doing other ones from my friend and i made the dumbest mistake.
like i expanded (x+1.1)^2 which became x^2+2.2x+1.21 and that was multiplied by 4.9 but i didnt multiply the last term. and i couldnt figure out why my answring was wrong :P
Untill i checked my work LOL