Asked by Vlad
A man stands on the roof of a building that is 50 m tall and throws a rock with a velocity of 35 m/s at an angle of 50 degrees above the horizontal
(a) What is the total time in the air
(b) What is the horizontal distance from the base of the building to the point where the rock strikes the ground
(c) What is the maximum height above the building
(d) What is the velocity of the rock just before it hit the ground
(a) What is the total time in the air
(b) What is the horizontal distance from the base of the building to the point where the rock strikes the ground
(c) What is the maximum height above the building
(d) What is the velocity of the rock just before it hit the ground
Answers
Answered by
Henry
h = 50 m.
Vo = 35m/s[50o]
Xo = 35*cos50 = 22.50 m/s.
Yo = 35*sin50 = 26.81 m/s.
a. Y = Yo + g*t = 0 at max ht.
26.81 - 9.8t = 0
9.8t = 26.81
Tr = 26.81/9.8 = 2.74 s. = Rise time.
hmax=ho + Yo*Tr - 0.5g*Tr^2
hmax=50+26.81*2.74 - 4.9*2.74^2=86.67 m.
Above gnd.
0.5g*t^2 = 86.67
4.9t^2 = 86.67
t^2 = 17.69
Tf = 4.21 s. = Fall time.
T = Tr+Tf = 2.74 + 4.21 = 6.95 s. = Time
in air.
b. D=Xo*T = 22.50m/s * 6.95s = 153.3 m.
c. h = Yo*Tr + 0.5g*Tr^2
h = 26.81*2.74 - 4.9*2.74^2 =
d. V^2 = 2g*hmax. Solve for V.
Vo = 35m/s[50o]
Xo = 35*cos50 = 22.50 m/s.
Yo = 35*sin50 = 26.81 m/s.
a. Y = Yo + g*t = 0 at max ht.
26.81 - 9.8t = 0
9.8t = 26.81
Tr = 26.81/9.8 = 2.74 s. = Rise time.
hmax=ho + Yo*Tr - 0.5g*Tr^2
hmax=50+26.81*2.74 - 4.9*2.74^2=86.67 m.
Above gnd.
0.5g*t^2 = 86.67
4.9t^2 = 86.67
t^2 = 17.69
Tf = 4.21 s. = Fall time.
T = Tr+Tf = 2.74 + 4.21 = 6.95 s. = Time
in air.
b. D=Xo*T = 22.50m/s * 6.95s = 153.3 m.
c. h = Yo*Tr + 0.5g*Tr^2
h = 26.81*2.74 - 4.9*2.74^2 =
d. V^2 = 2g*hmax. Solve for V.
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