a) A crazy New Yorker is running across the rooftops of buildings. He jumps from

Question

a) A crazy New Yorker is running across the rooftops of buildings. He jumps from roof to roof and usually makes it! In one case, he runs off a roof horizontally at 7.8 meters per second and hopes to land on the next building's roof, which is 2 meters lower than the roof he jumped from. How far can the separation distance between the two roofs be so that he doesn't fall between the building? b) How fast will his vertical speed be when he lands on the second roof? m/S c) How fast will his total speed be when he lands on the second roof? m/s

Explanation / Answer

a)

consider the motion along the Y-direction

Voy = initial velocity = 0 m/s

a = acceleration = 9.8

t = time of travel

Y = displacement = 2 m

using the equation

Y = Voy t + (0.5) a t2

2 = 0 t + (0.5) (9.8) t2

t = 0.64 sec

consider the motion along the X-direction

X = horizontal displacement

Vox = initial velocity = 7.8 m/s

t = 0.64 sec

using the equation

X = Vox t

X = 7.8 x 0.64

X = 5 m

b)

Voy = initial velocity = 0 m/s

a = acceleration = 9.8

Vfy = final velocity in vertical direction

t = time of travel = 0.64

using the equation

Vfy = Voy + at

Vfy = 0 + 0.64 x 9.8

Vfy = 6.3 m/s

c)

Vfy = 6.3 m/s

Vfx = Vox = 7.8 m/s

net speed is given as

Vf = sqrt(Vfx2 + Vfy2) = sqrt((7.8)2 + (6.3)2) = 10.03 m/s

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