Steven W. answered • 09/30/16
Tutor
4.9
(4,315)
Physics Ph.D., college instructor (calc- and algebra-based)
Hi Rae!
With these projectile motion problems, I like to organize them by first determining which kinematic quantity are we being asked to calculate. In this case, that is the height (effectively, the displacement) of the ball above the ground. This is a vertical direction quantity, and so I will work in the vertical direction first.
VERTICAL
------------
to find: displacement, Δy
To solve for one kinematic quantity with our usual set of kinematic equations, we have to know at least three of the other four. So let's determine which other vertical quantities we know:
know: acceleration (ay); initial velocity (voy)
ay = -9.8 m/s2 (we always know this right away in the vertical direction for a projectile on Earth; I am taking down to be negative)
voy = 0 (we know this because the only initial velocity off the racket is horizontal, of which no part is in the vertical; so, as far as vertical motion is concerned, it started from rest, as if it were dropped)
But this is pretty much the end of what we know. We are not told anything about final vertical velocity or time directly. So we are stuck in the vertical at this point.
Fortunately, one of the two quantities we can learn for the vertical is time (t), and time is the one kinematic quantity that crosses from the horizontal to the vertical direction in projectile motion. So we can use the horizontal direction to try to solve for time.
HORIZONTAL
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In the horizontal direction for a projectile, there is no acceleration (ax = 0), because the only acceleration for a projectile, by definition, is gravity, and gravity is only in the vertical direction. As a result of the acceleration being 0 horizontally, the horizontal velocity for a projectile is constant from the time it is launched to the time it lands (or hits something).
As a result of these facts, the kinematic equations in the horizontal direction for a projectile boil down to only one: the equation of motion for an object with constant velocity:
Δx = vxt
where
Δx = horizontal displacement
vx = (the constant) horizontal velocity
t = time of flight
So working with the horizontal direction for a projectile becomes very straightforward. There is one kinematic equation, and if we want to solve for one of those quantities, we have to know the other two.
In this case, we want to solve for time (t). Fortunately, we are told both the (constant) horizontal velocity (vx = 28.1 m/s) and the horizontal displacement of the ball in flight (Δx = 19.8 m). So we have:
Δx = vxt
19.8 m = (28.1 m/s) t
t = 19.8/28.1 = 0.70 s
*****************************************************
Now, we can carry this time back to the vertical direction, and we have our third known to solve for Δy.
to find: Δy
know: ay (=-9.8 m/s2), voy (= 0 m/s; starts from rest vertically), t (=0.70 s)
Then we can use the kinematic equation that connects these four quantities to solve for Δy. That equation is:
Δy = voyt + (1/2)ayt2 = (0 m/s)(0.70 s) + (1/2)(-9.8 m/s2)(0.70 s)2 = 0 - 2.40 m = -2.40 m
Now, the displacement vertically is negative because I defined down as negative, and the ball ended up lower than where it started vertically. However, it is hopefully a short intuitive leap to realize this means the ball started 2.40 m above the ground (a little over 6 ft, not an unreasonable height to hit a tennis ball).
I hope this helps! I will check again later to see if I made any math errors, but this procedure of choosing a direction, determining what to calculate, then determining what you know, should generally work well for these problems. If you have any other questions about this, just let me know!
