A small rock is launched from the ground with an initial velocity of magnitude v

Question

A small rock is launched from the ground with an initial velocity of magnitude vo = 16.0 m/s and diriection theta = 53.1 degrees above the horizontal. The rock strikes a tall building 2.20 s after being launched. Neglect air resistance and assume the ground is level.

A) What is the horizontal distance from the point where the rock is launched to the building?

B) At what vertical height above ground does the rock strike the building?

C) What is the speed of the rock just before it strikes the building?

Explanation / Answer

verticle velocity = 16*sin(53.1) = 12.8 m/s

horizontal velocity = 16*cos(53.1) = 9.6 m/s

A)

using equation

D = ut + (1/2)at^2

D = 9.6*2.2 + 0 [a = 0]

D = 21.12 m [horizontal distance]

B)

D = 12.8*2.2 - (1/2)*9.8*2.2^2 = 4.44 m [verticle height]

C)

using eqation for verticle velocity

v^2 = u^2 + 2*a*s

v^2 = (12.8)^2 - 2*9.8*4.44

v = 8.76 m/s [verticle velocity]

velocity of rock just before it strikes the building = sqrt(8.76^2 + 9.6^2) = 13 m/s

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