Question

My question is whats the velocity i need to throw a dart at the dart board to hit the bulls eye if my initial height is level with the bulls eye 1.7m is the height and 2.4m is the distance away from the board. can you show the formula you used and the values you subed then in for me

Answer 1

can u explain what is 1.7m the height.

Answer 2

1.7m is the height from the center of the bulls eye

Answer 3

to the ground

Answer 4

And how high up are you throwing the dart from?

Answer 5

2.4m is the distance away

Answer 6

It is projectile motion. And many combinations of speed and angle are possible.

Answer 7

well lets say in a controlled environment no air friction, drag just gravity

Answer 8

Ah, I see. So you throw it level to the bulls-eye. So you're throwing from 1.7m high, too.
So koushik_ksv is right with his statement. And the math can show that, with some work!

Answer 9

i was think about it a little would i use \[y = \frac{ 1 }{ 2 }at^2 + v_{0}t+y_{0}\] to solve this

Answer 10

formula for projectile motion is:

\[y = x*\tan(\alpha) - \frac{ g*x^2 }{ 2*v0 } * (1 + tg^2\alpha)\]

but you have two unknown variables, alpha (angle at which you are throwing the dart) and v0, which is initial velocity

Answer 11

sorry, v0 is squared

Answer 12

initial velocity would be zero

Answer 13

initial velocity is what you are looking for in this assignment, if I understood it correctly. by that, I mean velocity you need for dart to have to reach it's target.

however, that velocity depends on angle at which you are throwing that dart

Answer 14

Example of different throws to make the bulls-eye. They change in angle an velocity.