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 Question from Dillon, a student: Hi, my name is Dillon and I need assisstance on a parametric word problem. "Ron throws a ball straight up with an initial speed of 60 feet per second from a height of 5 feet. Find parametric equations that describe the motion of the ball as a function of time. How long is the ball in the air? When is the ball at its maximum height? What is the maximum height of the ball?"

Hi Dillon.

The standard equations of projectile motion are

 y(y0, v0, t, θ) = y0 + v0 t sinθ - ½ g t2. and x(x0, v0, t, θ) = x0 + v0 t cosθ.

(Sometimes it is written as + ½ instead of - ½, but then you need to ensure g is a negative number).

Where:

• y() is the vertical position (height) and x() is the horizontal position
• x0 and y0 are the initial horizontal and vertical positions the projectile is launched from
• v0 is the initial velocity of the projectile
• θ is the angle of the initial trajectory with the horizontal (i.e.: relative to the ground)
• t is the time in seconds since the launch and
• g is the acceleration due to gravity (a fixed amount of 32 ft/sec2, at least on earth).

So in your question, y0 is 5 ft, v0 is 60 ft/sec, θ is 90°. So this equation simplifies to:

y( t ) = 5 + 60t - 16 t2.

(Remember that sin 90° is 1, so that whole term can be removed).

1. How long is the ball in the air?

That would be the value of t for which the height is 0 ft above the ground.

0 = 5 + 60t - 16 t2. Solve for t.

2. When is the ball at its maximum height?

Notice that this equation is a parabola. That means that if y is the vertical axis and t is the horizontal axis, that we have a parabola opening downwards. The question asks, where is the vertex of this parabola? What is the horizontal (y) value at that point? If you put the equation into standard form for a parabola, you can easily read off the vertex.

3. What is the maximum height of the ball?

You pretty much already solved this if you finished question 2. This is just the vertical coordinate of the vertex.

Hope this helps,
Stephen La Rocque.

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