Posted by admin | Posted in archery | Posted on 02-01-2011
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A 0.59 kg rock is pro jected from the edge of the top of a building with an initial velocity of 7.45 m/s at?
A 0.59 kg rock is pro jected from the edge of
the top of a building with an initial velocity of
7.45 m/s at an angle 44 ◦
above the horizontal.
The building is 10.9 m in height.
At what horizontal distance, x, from the
base of the building will the rock strike the
ground? Assume the ground is level and
that the side of the building is vertical. The
acceleration of gravity is 9.8 m/s
2
.
Answer in units of m
The vertical component of velocity[Uy] = usinθ = 7.45 x sin44* = 5.18 m/s
The horizontal component of velocity[Ux] = ucosθ = 7.45 x cos44* = 5.36 m/s
Let the rock attain the maximum height (h) in t1 sec,
=>By v = u - gt
=>0 = [Uy] - 9.8t1
=>t1 = 5.18/9.8 = 0.53 sec
By s = ut - 1/2gt^2
=>h = 5.18 x 0.53 - 1/2 x 9.8 x (0.53)^2
=>h = 1.38 m
Thus the total height from ground(H) = h + 10.9 = 12.28 m
Let the rock take t2 sec to fall H meter
=>By s = ut + 1/2gt^2
=>12.28 = 0 + 1/2 x 9.8 x t2^2
=>t2 = √2.51
=>t2 = 1.58 sec
Total time of flight(t) = t1 + t2 = 0.53 + 1.58 = 3.16 sec
Thus the horizontal distance(R) covered by rock:-
=>By R = [Ux] x t
=>R = 5.36 x 3.16 = 16.95 m
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