A dummy is fired vertically upward from a cannon with a speed of 40 m/s. How long is the dummy in the air? What is the dummy's maximum height? sdsdsdsdsdsdsdsdsdsd
A dummy is fired vertically upward from a cannon with a speed of 40 m/s. How long is the dummy in the air? What is the dummy's maximum height?
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If you recognize that the graph of height versus time will be a parabola (here air resistance is being ignored, otherwise the graph would be more complex), the symmetry of the curve requires that the slope at time 0 will have the same magnitude and opposite sign as the slope when the dummy returns to the ground. Meaning at impact v = - 40 m/s.
In fact, the velocity is a linear function of time with its slope equal to the acceleration.
v = -9.8 m/s2 * t + 40 m/s
Again the dummy impacts the ground with a velocity of -40 m/s, so you must substitute this value for v and solve for the time of impact to learn how long the dummy was in the air.
-40 m/s = -9.8 m/s2 * tImpact + 40 m/s
-40 m/s - 40 m/s = -9.8 m/s2 * tImpact + 40 m/s - 40 m/s
-80 m/s / (-9.8 m/s2) = -9.8 m/s2 * tImpact / (-9.8 m/s2)
tImpact ≈ 8.163 s (If you pay attention to significant digits, this value should be rounded to 8 s since 40 m/s has only one significant digit, but most text books are not careful and will report the answer as 8.2 s or 8.16 s)
To find the maximum height, recognize that symmetry of the curve requires that maximum (vertex of the parabola) height occur half way between the time the object left the ground and the time it struck the ground.
tMax Height = (1/2) * tImpact
tMax Height ≈ 4.0816 s (Note that I used the exact time in the calculator rather than the rounded time to calculate this value. You can round the value if the instructor wants this time, but in this problem, the time at the top is not requested. Any rounding will introduce error into your calculation of the maximum height.)
Due to the symmetry of the curve, this is also the amount of time required for the dummy to fall from the maximum height to the ground.
The equation for the position of an object undergoing constant acceleration describes the height (y) of the dummy as a function of time.
y = (1/2) a * t2 + v0 * t + y0
In this case the equation becomes:
y = (1/2) (-9.8 m/s2) * t2 + 40 m/s * t + 0 m
y = (-4.9 m/s2) * t2 + 40 m/s * t
(An alternative approach to solving the problem [or checking the answer] with a graphing calculator is to replace the t with x and graph the function. Have the calculator fined the maximum and the x-intercept.)
Substitute the value of tMax Height for t in the equation above to determine the maximum height the dummy reached.
yMax ≈ 81.63 m
Round this value as required. I prefer to write values with only one significant digit using scientific notation. (e.g. 8. × 101 m/s)
The formula for the height of an object fired straight up is:
h = (1/2)gt2 + v0t + h0
Where h is the height at any time t, g is the acceleration due to gravity, v0 is the initial velocity, and h0 is the initial height. We know:
g = -9.8 m/s2
v0 = 40 m/s
h0 = 0 m (we start at the ground)
How long is the dummy in the air?
The dummy will be in the air until h=0 again. So:
0 = (1/2)gt2 + v0t + h0
Use the quadratic formula to find t:
t = (-v0/g) ± (1/g)√(v02-gh0)
Plugging in the values for g, v0, and h0, we get:
t = (4.08) ± (4.08)
t = 0, 8.16 sec
t=0 represents the fact that the dummy starts out on the ground (h=0), so t = 8.16 represents the dummy landing back on the ground after being fired. So our answer for how long the dummy is in the air is:
t = 8.16 seconds
What is the dummy's maximum height?
The graph of the equation for the height of the dummy, h = (1/2)gt2 + v0t + h0, is a parabola. Since the coefficient of the t2 term, (1/2)(-9.8), is negative, it is an upside down parabola with the vertex at the top. The vertex thus represents the highest point on the parabola. The t coordinate of the vertex is -b/2a = (-40)/(-9.8) = 4.08 seconds. Plug that value of t into the height equation to find the maximum height of the dummy.