just do the vertical problem
Vi = 22.5 sin 19.3 = 7.44 m/s upwards
v = Vi - 9.8 tup
at top v = 0 so
tup = 7.44/9.8 = .759 seconds upward
so how long did it fall from the zero vertical velocity high spot?
t = 3.50 - .759 = 2.74 seconds falling
so how high was it at peak?
fall dist = 4.9 t^2 = 4.9(2.74)^2
= 36.8 meters from peak of trajectory
now subtract the roof to peak of flight
36.8 = roof + (7.44/2)(.759)
36.8 = roof + 2.82
roof = 34 meters
A brick is thrown upward from the top of a building at an angle of 19.3◦ above the horizontal and with an initial speed of 22.5 m/s. The acceleration of gravity is 9.8 m/s2 . If the brick is in flight for 3.5 s, how tall is the building? Answer in units of m
4 answers
Vo = 22.5m/s[19.3o].
Yo = 22.5*sin19.3 = 7.44 m/s = Ver. component of initial velocity.
Y = Yo + g*Tr = 0 at max. ht. =
7.44 + (-9.8)Tr = 0,
Tr = 0.759 s. = Rise time.
T = 2*Tr = 2 * 0.759 = 1.52s = Time to return to top of roof.
T+Tf = 3.5s.
1.52 + T
Yo = 22.5*sin19.3 = 7.44 m/s = Ver. component of initial velocity.
Y = Yo + g*Tr = 0 at max. ht. =
7.44 + (-9.8)Tr = 0,
Tr = 0.759 s. = Rise time.
T = 2*Tr = 2 * 0.759 = 1.52s = Time to return to top of roof.
T+Tf = 3.5s.
1.52 + T
Vo = 22.5m/s[19.3o].
Yo = 22.5*sin19.3 = 7.44 m/s = Ver. component of initial velocity.
Y = Yo + g*Tr = 0 at max. ht. =
7.44 + (-9.8)Tr = 0,
Tr = 0.759 s. = Rise time.
T = 2*Tr = 2 * 0.759 = 1.52s = Time to return to top of roof.
T + Tf = 3.5s.
1.52 + Tf = 3.5,
Tf = 1.98s = Fall time from roof to gnd.
h = Yo*Tf + 0.5g*Tf^2.
h = 7.44*1.98 + 4.9*1.98^2 = 34 m. = Ht. of bldg.
Yo = 22.5*sin19.3 = 7.44 m/s = Ver. component of initial velocity.
Y = Yo + g*Tr = 0 at max. ht. =
7.44 + (-9.8)Tr = 0,
Tr = 0.759 s. = Rise time.
T = 2*Tr = 2 * 0.759 = 1.52s = Time to return to top of roof.
T + Tf = 3.5s.
1.52 + Tf = 3.5,
Tf = 1.98s = Fall time from roof to gnd.
h = Yo*Tf + 0.5g*Tf^2.
h = 7.44*1.98 + 4.9*1.98^2 = 34 m. = Ht. of bldg.
My first post was done in error. Please disregard it.
0 answers