in response to the school shutdown, my teacher had given me rather difficult work that i cant seem to even understand
many of you may be unaware that mild the mild-mannered Cameron is in fact the superhero "El maestro." my first clue to her identity was her superhuman ability to create math and physics comics; another one of her gift if her ability to jump very high. i recently witnessed her leap upwards off the roof of a building at 38.9 m/s. i also noted that "El maestro" landed on the ground 8.54 seconds after she left the roof. calculate
a) the maximum height above the roof reached by "El maestro"
b)her impact velocity (final velocity?) with the ground
c)the height of the roof measured from the ground (initial velocity?)
*fine print*
warning! do not try this at home remember Cameron is "El maestro" and you are not
h(t) = h0 + v0*t - 4.9t^2
So, you have
h0 + 38.9*8.54 - 4.9*8.54^2 = 0
That gives you the initial height.
v = v0 - 9.8t
the max height is attained at t = -b/2a = 38.9/9.8
a. V^2 = Vo^2 + 2g*h = 0
38.9^2 + (-19.6)h = 0
h = 77.2 m.
b. V = Vo + g*Tr = 0
38.9 + (-9.8)Tr = 0
Tr = 3.97 s. = Rise time.
3.97 + Tf = 8.54
Tf = 4.57 s. = Fall time.
h = 0.5g*Tf^2 = 4.9*4.57^2 = 102.3 m. above gnd.
V^2 = Vo^2 + 2g*h = 0 + 19.6*102.3 = 2005.8
V = 44.8 m/s.
c. 77.2 + h = 102.3
h =
To solve this physics problem, we can use the equations of motion to find the answers. Let's break down the problem into parts:
a) To determine the maximum height reached by "El maestro," we can use the equation:
v² = u² + 2as,
where:
v = final velocity (0 m/s at the maximum height),
u = initial velocity (38.9 m/s),
a = acceleration (gravity, approximately -9.8 m/s²),
s = displacement (height).
Rearranging the equation, we get:
0² = (38.9)² + 2(-9.8)s.
Simplifying further:
0 = 1509.21 - 19.6s.
Now, solving for s:
19.6s = 1509.21,
s = 1509.21 / 19.6.
s ≈ 77.05 meters.
Therefore, "El maestro" reached a maximum height of approximately 77.05 meters above the roof.
b) To find the impact velocity or final velocity with the ground, we can use the equation:
v = u + at,
where:
v = final velocity,
u = initial velocity (0 m/s),
a = acceleration (gravity, approximately -9.8 m/s²),
t = time (8.54 seconds).
Plugging in the values, we have:
v = 0 + (-9.8)(8.54).
v ≈ -83.692 m/s.
The negative sign indicates that the velocity is directed downwards. So, the impact velocity of "El maestro" with the ground is approximately 83.692 m/s downwards.
c) Lastly, to determine the height of the roof measured from the ground (initial velocity), we can use the equation:
s = ut + (1/2)at².
Rearranging the equation, we have:
s = ut + (1/2)at²,
0 = u(8.54) + (1/2)(-9.8)(8.54)².
Simplifying further:
0 = 8.54u - 0.5(9.8)(73.1716).
Now solving for u:
390.5456 = 8.54u,
u = 390.5456 / 8.54.
u ≈ 45.763 m/s.
Therefore, the height of the roof measured from the ground (initial velocity) is approximately 45.763 meters per second.
Remember, it's just a theoretical calculation, so please don't try to replicate it in real life.