Projectile Motion: Solving for Time and Initial Velocity

[Unsettledchim]

Homework Statement

A cannonball is fired from the ground at an angle of 50 degrees. The ball lands 67 meters from where it was fired. Air resistance is negligible.

How long was the ball in the air?

With what velocity was the cannonball fired?

Homework Equations

X_f=.5at²+V_ot+X_o
V_f=at+V_o

The Attempt at a Solution

 

[Cryxic]

Homework Statement

A cannonball is fired from the ground at an angle of 50 degrees. The ball lands 67 meters from where it was fired. Air resistance is negligible.

How long was the ball in the air?

With what velocity was the cannonball fired?

Homework Equations

X_f=.5at²+V_ot+X_o
V_f=at+V_o

The Attempt at a Solution

You will need the range equation:

R = v^2 * sin(2*theta) / g. Knowing theta and R (the range), solve for v.

The time in the air is given by t = 2*v*sin(theta)/g. You know v and theta, so solve for t. If you need me to derive these equations, let me know.

 

[Unsettledchim]

Thank you and yes that would be great if you could drive the equation

 

[Cryxic]

Thank you and yes that would be great if you could drive the equation

The total velocity can be written as components of the velocity in the x direction and in the y direction:

V(total) = sqrt(V(x)^2 + V(y)^2)

where V(x) = V(0)cos(theta) and V(y) = V(0)sin(theta)

In the x direction, the object travels the following distance (let's call it D(x), which is actually the range R in my last post):

D(x) = V(x)*t = V(0)cos(theta)*t

Our goal is now to find an expression for t, the time, and we can do that by examining the motion in the y direction:

D(y) = V(y)*t - (1/2)*g*t^2 = V(0)sin(theta)*t - (1/2)*g*t^2

Now take the derivative with respect to time and set it equal to zero. This will tell you the time at which the object's velocity in the y direction is zero. Practically, you're finding out when the object reaches the top of the parabolic curve (the greatest height in its motion).

D'(y) = V(0)sin(theta) - g*t = 0

Solve for t:

t = V(0)sin(theta) / g.

The time to complete the entire distance is obviously twice this:

t = 2V(0)sin(theta)/g.

Plug this t back into the D(x) equation to get:

D(x) = 2*V(0)^2*sin(theta)*cos(theta)/g

You can leave the equation in this form if you want, but making the trig substitution sin(2*theta) = 2sin(theta)cos(theta) makes it look nicer:

D(x) = V(0)^2sin(2*theta)/g

 

[Unsettledchim]

Thank you, that helped a ton, my physics teacher had not given us that equation.