Projectile Motion - Finding Ini. Vel. using Range and Angle

[Tearsandrille]

Homework Statement

A quarterback throws the football to a stationary receiver who is 31.5m down the field. If the football is thrown at an initial angle of 40.0 degrees to the ground, at what initial speed must the quarterback throw the ball for it to reach the receiver, if the ball is caught at the same level it was thrown?

Homework Equations

R = (Vo²*sin(2*theta))/g

The Attempt at a Solution

The issue I am having is that if gravity is negative I don't get a real number, but I can not understand or explain why gravity would be positive.

 

[tiny-tim]

Hi Tearsandrille!

(have a theta: θ )

R = (Vo²*sin(2*theta))/g

Did you just copy that equation from a book?

Yes it's correct, but you'll never be able to remember equations like that in the exam, and anyway you need to be able to prove them yourself …

solve the simultaneous equations for the x and y directions (eliminating t), and you'll see why it doesn't matter that g is negative.

 

[Tearsandrille]

No, I did not just copy the equation from the book. I did derive it using the fact that R = Vxinitial*t and t = (2Vosin θ)/g (which I derived from t = (Vf - Vi)/-g)

So, the reason that I don't need to worry about the negative when I use the range equation is that I already compensated for that when I solved for t in the equation Vf=Vi + gt.

Thank you.

 

[tiny-tim]

So, the reason that I don't need to worry about the negative when I use the range equation is that I already compensated for that when I solved for t in the equation Vf=Vi + gt.

Yup!

 

[rwisz]

As a scientist, it is important to understand the context in which the equations are used. In this case, the equation you have provided is for the horizontal range of a projectile, assuming that the initial velocity is perpendicular to the ground and there is no air resistance. In this case, gravity is acting in the vertical direction, which is why it is negative in the equation.

To solve this problem, we can use the equation for the vertical displacement of a projectile:

y = yo + vo*t - (1/2)gt^2

Where:
y is the vertical displacement (in this case, it is 0 since the ball is caught at the same level it was thrown)
yo is the initial vertical position (in this case, it is also 0)
vo is the initial vertical velocity (we are solving for this)
t is the time it takes for the ball to reach the receiver (we can find this using the horizontal range and the initial horizontal velocity)

We can rearrange the equation to solve for vo:

vo = (R*sin(theta))/t + (1/2)gt

Substituting in the given values:
R = 31.5m
theta = 40 degrees
t = R/(vo*cos(theta))

Plugging these values into the equation, we get:

vo = (31.5m*sin(40 degrees))/((31.5m)/(vo*cos(40 degrees))) + (1/2)(9.8m/s^2)(31.5m)/(vo*cos(40 degrees))

Simplifying, we get:

vo = 16.8m/s

Therefore, the initial speed at which the quarterback must throw the ball is 16.8m/s. It is important to note that this is the magnitude of the initial velocity, and the direction will depend on the angle at which the ball is thrown.