Projectile Motion Help: Angle 15, Time 3.5s, Unknown Distances & Velocity

In summary: Ah, well, putting y_0 = 0 then you can solve:You can solve the parabolic equation y = x \tan(\theta _o) - \frac{gx^2}{2v_o \cos^2(\theta _o)}for the initial velocity v_o - it is the only unknown in the equation if you use the other equations to subs for x and y.But the problem do not require the initial velocity to be confined by any of the given values. This means that even if the projectile is resting on the launching pad at fifteen degrees and time marches on for 3.5 seconds the conditions are still met! So one can shoot the projectile at any initial speed and the situation
  • #1
squall325
2
0
1. Angle = 15, Time(s) = 3.5, Horizontal Distance(m) = ?, Vertical Distance(m) = ?, Initial Velocity(m/s) = ?, Final Velocity(m/s) = ?



2. Our teachers told us to use y = vi(sine(theta)(t) - (gt^2)/2 - but we can't get an answer with only 2 givens.



3. Cant get an answer using the given equation. Hope you guys could help.
 
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  • #2
Indeed, you can't solve for so many unknowns.
 
  • #3
Matthaeus_ said:
Indeed, you can't solve for so many unknowns.

Sure you can, if you assume 'time' is total time of flight and the surface is horizontal (etc etc). y=0 at t=0 and y=0 at t=3.5 sec. Use that to solve for vi and vertical distance. Now use x=vi*cos(theta)*t to do the horizontal part.
 
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  • #4
Maybe you are to assume the object is to be thrown from ground level? Then this is solvable.
 
  • #5
Maybe you are to assume the object is to be thrown from ground level? Then this is solvable.
 
  • #6
Ah, well, putting [tex]y_0 = 0[/tex] then you can solve:

[tex]\displaystyle \left \{ \begin{array}{ll}
y &= (v_0 \sin 15^{\circ})t - \frac{1}{2} gt^2\\
v_y &= (v_0 \sin 15^{\circ}) - gt\\
x &= (v_0 \cos 15^{\circ})t
\end{array}[/tex]
 
  • #7
You can solve the parabolic equation

[tex]y = x \tan(\theta _o) - \frac{gx^2}{2v_o \cos^2(\theta _o)}[/tex]

for the initial velocity [tex]v_o[/tex] - it is the only unknown in the equation if you use the other equations to subs for x and y.

But the problem do not require the initial velocity to be confined by any of the given values. This means that even if the projectile is resting on the launching pad at fifteen degrees and time marches on for 3.5 seconds the conditions are still met! So one can shoot the projectile at any initial speed and the situation will still be within limits of the given data.
 
Last edited:

1. What is projectile motion?

Projectile motion refers to the motion of an object that is launched into the air and moves under the force of gravity. This type of motion is influenced by both the initial velocity and the angle at which the object is launched.

2. How do I calculate the initial velocity of a projectile?

The initial velocity of a projectile can be calculated using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s^2), and t is the time. Alternatively, you can use the formula v = u*sin(theta), where theta is the launch angle.

3. What is the range of a projectile?

The range of a projectile is the horizontal distance it travels before hitting the ground. It can be calculated using the formula R = (u^2*sin(2*theta))/g, where u is the initial velocity and theta is the launch angle.

4. How does air resistance affect projectile motion?

Air resistance, also known as drag, can affect projectile motion by slowing down the object and reducing its range. This is because air resistance acts in the opposite direction of the object's motion and increases as the object's speed increases.

5. How can I use projectile motion in real life?

Projectile motion is used in various real-life scenarios, such as sports (e.g. throwing a ball or shooting a basketball), fireworks displays, and launching objects from a catapult. It is also important in engineering and physics experiments to understand the effects of gravity and air resistance on moving objects.

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