# Find angle (Projectile Motions)

#### Stingarov

Alright so I thought I understood this well.

I'll give an example of how I solve for range and such with the given information:

A ball is thrown from a height of 1.86 m, at a speed of 21 m/s and at an angle of elevation of 59.4o. How far from the person who threw it does the ball land?

Range: Vxt
Vx: Vcos Theta
Vy: V sin Theta

The method in which I solve for t goes as follows:

Vy = 21 * sin59.4

Vy = 18.07

Maximum Height: v²= 2a(s) +existing height of 1.86

Maximum height: 18.07² /2(9.8) = s -> 16.66 + 1.86 = 18.52m

Total time: Time down + Time Up
s + 1.86 = 1/2at² + s = 1/2at²

18.52*2 / 9.8 = t² + 16.66 * 2/9.8 = t²

1.94 + 1.84 = 3.78s

R = V cos Theta * t
R = ~40.4 m

The problem I encounter is simply that I do not know how, using this method, or by any other really, to compute the angle or initial velocity if Total time is not a given already.

Say the equation was rearranged and said:

A person threw the ball at an initial rate of 21 m/s from a height 1.86. It hit the ground at 40.4m. What was the angle of projection?

Can anyone help show me how to work these backwards? Would be highly appreciated.

#### enricfemi

A person threw the ball at an initial rate of 21 m/s from a height 1.86. It hit the ground at 40.4m. What was the angle of projection?
v=21
h=1.86
(a means angle)
Vx=V*cos(a)
Vy=V*sin(a)
when the ball hit ground ,x=40.4,y=0.
put it into the equation andrevdh gave to you:
Vx*t=x
$$y = v_{yo}t - \frac{1}{2}gt^2$$

you may solve it,though it may difficult in maths.

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#### andrevdh

Homework Helper
For the given equation y = ....
y will not be zero when the ball lands!

#### enricfemi

For the given equation y = ....
y will not be zero when the ball lands!
Oh,Yes,I am so sorry!

#### andrevdh

Homework Helper
It seems you are trying to solve a though problem.
See the section :"Angle θ required to hit coordinate (x,y)" in Wikipedia:

http://en.wikipedia.org/wiki/Trajectory_of_a_projectile" [Broken]

Last edited by a moderator:

#### enricfemi

Like this y=h+Vy*t-(1/2)*g*t*t

#### andrevdh

Homework Helper
close, more like

$$y = y_o + v_{yo}t - \frac{1}{2}gt^2$$

or if you take the launching point as (0,0) y = - 1.86

#### Stingarov

Ah alright, thanks

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