Question

A projectile is fired with initial velocity of v0 feet per second. The projectile can be pictured as being fired from the origin into the first quadrant, making an angle (theta) with the positive x-axis, as shown in the figure. If there is no air resistance, then at t seconds the coordinates of the projectile (in feet) are x = (v0t)(cos(theta)) and y = -16t^2 + (v0t)(sin(theta)). Suppose a projectile leaves the gun at 100 ft/sec and theta = 60.

f) show that y = ((-16sec^2(theta))/v0^2)(x^2) + tan(theta)

Answer 1

If you examine the final expression for y, you will find that "t" is missing.
So to achieve this, you would solve for t using the first equation, and substitute t in the second equation to arrive at the final results.
I'll start the first step:
\(x = (v_0~t)(\cos(\theta))\)
solve for t:
\(t=x/(v_0~\cos(\theta))\)..................(1)
\( y = -16t^2 + (v_0~t)(\sin(\theta))\)..........(2)
So substitute (1) into the second equation (2) wherever you see "t", and simplify to get the required expression for y.

Answer 2

I simplified it to y=16(x/v0cos(theta))^2+(v0(x/v0cos(theta))(sin(theta))(v0), but I do not know how I can go about to simplify the rest

Answer 3

Hints:
1/cos(theta)=sec(theta)
so
1/cos^2(theta)=sec^2(theta)

in the second term, you see
v0/v0 somewhere, they can be cancelled, but I think you have a v0 too many somewhere.
Also the second term of the identity is missing an x.
I think it should read:
f) show that y = ((-16sec^2(theta))/v0^2)(x^2) + \(\color{red}{x}\)tan(theta)

Answer 4

If you still don't get it, post your calculations with all the steps, and we'll help you figure out what went wrong.

Answer 5

Thank you!

Answer 6

You're welcome! :)