Question

Projectile Motion Question! Please help.

Answer 1

Can any of you make sense of this please?

If something is launched at 52.3 m/s at ground level at an angle of 35 degrees above the ground. Use the equation to determine the horizontal displacement of the object.

Would Vi=52.3
Vf=35
g=9.81?

Answer 2

\[\Delta x = \frac{ v_i^2 }{ g } \sin(2\theta) \implies \frac{ (52.3)^2 }{ 9.81 }\sin(2*35)\]

Answer 3

@QuantumMechanics where did you get sin from?

Answer 4

They want you to use the equation correct?

Answer 5

I was told that I can use the UAM equations to solve this problem.

Answer 6

Wait it wants me to use the equation provided

Answer 7

You may. but just use the equation they gave you, it's the range formula, that gives the horizontal displacement. Look at your uploaded image

Answer 8

lol @QuantumMechanics I just realized how stupid my question was, sorry for wasting your time and I really appreciated your help! But where did you get the two from in the equation set up?

Answer 9

It's from the formula, just look at your attachment

Answer 10

Yeah I just realized that lol, I am so sorry @QuantumMechanics

Answer 11

No worries, take care!

Answer 12

@QuantumMechanics come back! okay I think I am going to need your guidance. Okay so I set it up as (52.3)2/9.81 (2*35) correct. I then squared 52.3 and got 2735.29 now should I divide it by 9.81,once I do that I will end up with 278.79 and then I should multiple 2*35 which will equal 70. But the answer I got looks odd,mind telling me where I went wrong?

Answer 13

If anyone is looking you are more than welcome to provide an input!!!

Answer 14

http://www.wolframalpha.com/input/?i=+%5Cfrac%7B+%2852.3%29%5E2+%7D%7B+9.81+%7D%5Csin%282*35%29

Answer 15

According to the answer form,the answer is 240. How sway? Either the wrong answer was entered or I am doing something wrong.

Answer 16

\[\frac{ (52.3)^2 }{ 9.81 } \sin(2*35) = 262m\]