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.

From now on,I advise you to learn more algbra.It is important in physics.

 

[andrevdh]

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!
It should add h in.

 

[andrevdh]

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"

 

[enricfemi]

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

 

[andrevdh]

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