A Projectile Motion Problem

Question

I need more help, but this time it if for a problem involving projectile motion.

The question is " A shot leaves the thrower’s hand 6.5 ft above the ground at a 45° angle at 44 ft / sec. Where is it 3 sec later?"

I tried substituting known values into the equation for a vector describing projectile motion, and ended up with r(t)= <44cos(45)t + [6.5 + (44sin(45) - 16t^2]>. I ended up with <93.338, -106.387> for the coordinates, and I tried graphing the vector on my graphing calculator.

On the other hand, the solution to the problem is "Shot put is on the ground, about 66 ft 3 in. from the stopboard." I tried solving the problem a second time and tried to parametrize the position vector to get x= 44cos(45) and y= 6.5 - 32t^2. I'm just lost now.

Answer 1

The fact that in your solution $y \lt 0$ says the shot ends up below the ground, which is not physically possible. You are expected to realize that and assume that the shot stops when it hits the ground. Solve $y(t)=0$ to get $t$, then plug in that value of $t$ to get $x$.

Answer 2

I would start by splitting the initial velocity vector into the horizontal and vertical directions. The initial velocity of $44$ ft/s has an angle of $45^\circ$, so...

• Initial horizontal velocity = $44*cos(45)\approx31.113$ ft/s
• Initial vertical velocity = $44*sin(45)\approx31.113$ ft/s

Now that we have the initial velocity vectors, we need to write equations that model the change in those velocities over time. If we disregard air resistance, then the horizontal vector will not have any other force acting on it, so horizontal velocity does not change.

Horizontal velocity = $31.113$ ft/s

However, the vertical velocity will change because the force of gravity will act against the upward motion of the projectile. The constant of gravitational acceleration is approximately $32.17$ ft/s$^2$. This means the the vertical velocity will decrease by $32.17$ ft/s every second. Therefore, the equation would be as following:

Vertical velocity = $(31.113 - 32.17t)$ ft/s

Now, we can use these equations to figure out how far the projectile traveled in $3$ seconds.

Horizontal position: $31.113*3\approx93.34$ ft

Calculating the vertical position is a little harder since the vertical velocity is constantly changing. So what value do we use here? We need the average vertical velocity over the time interval, because the average is representative of all the velocity values that occur during that time interval. Conveniently, the equation for vertical velocity is a linear function, so we can find the average value of the function simply by adding the initial and final values in the time interval, then dividing by $2$.

Initial vertical velocity = $31.113$ ft/s

Final vertical velocity after $3$ seconds = $(31.113 - 32.17(3))=-65.397$ ft/s

Average vertical velocity over $3$ seconds = $\frac{31.113-65.397}{2}=-17.142$ ft/s

Now, we can plug this into the distance formula.

Vertical position: $-17.142*3=-51.426$ ft

This negative vertical position value means that the projectile has gone $51.426$ feet lower than its starting height of $6.5$ feet. This implies that the projectile has gone underground, but since that's impossible, we can assume that the projectile hit the ground and stayed at ground level. According to the actual solution (which you mentioned in the question body), it seems that the projectile completely stopped moving once it hit the ground, since $66.25$ feet is less than the distance of $93.34$ feet, the distance it would have traveled if it had maintained its horizontal motion for the entirety of $3$ seconds. Therefore, we now have to figure out exactly how long it took for the projectile to hit the ground, and use that time to calculate how far forward it traveled before hitting the ground and stopping.

For vertical distance, we know to use the average vertical velocity since the instantaneous value is always changing.

Average vertical velocity = $\frac{31.113+(31.113 - 32.17t)}{2}$

Now for vertical distance, we multiply the average velocity by time, then add $6.5$, which was its starting height.

Vertical distance = $\frac{31.113+(31.113 - 32.17t)}{2}t+6.5$

Now we equate this to zero, and solve for $t$, which tells us how long the projectile was in the air before hitting the ground and stopping.

$\frac{31.113+(31.113 - 32.17t)}{2}t+6.5=0$

$t=1.124$ seconds

Plug this into the horizontal distance formula:

$31.113*1.124\approx66.084$ ft

This means that the projectile traveled $66.084$ feet forward before hitting the ground and immediately stopping. This is pretty close to the actual answer, and I'm sure it would have been exact if I hadn't done so many approximations.