* PLEASE answer no.5 and 6 with FULL WORKINGS to your answer. Thanks* 1.) How mu
ID: 1463115 • Letter: #
Question
* PLEASE answer no.5 and 6 with FULL WORKINGS
to your answer. Thanks*
1.) How much time does it take the object to reach the highest point in its trajectory?
2.) how much time would it take the object to travel a horizontal distance of 6.00m from its launch point?
3.)at the instant when the object has traveled a horizontal distance of 6.00m from its launch point, the scalar y component of the object's velocity would be?
4.)At what instant when the object has traveled a horizontal distance of 6.00m from its launch point, the object's speed would be?
5.)The object would pass over the near side of building B
a. 4.0m above the level of building B's roof.
b. 0.65m above the level of building B's roof.
c. 0.43m above the level of building B's roof.
d. 0.34m above the level of building B's roof.
6.) The object would land on the roof of building B
a. about 0.79m beyond the near side of building B.
b. 6.47m beyond the near side of building B
c. the object will fall short and land somewhere down the alley
d. about 6.79m beyond the near side of building B
e. about 0.47m beyond the near side of building B
* please answer no.5 and 6 with clear workings to your answer. Thanks*
Explanation / Answer
along horizantal
x = vox*T
T = x/vox = x/(vi*cos60)
along vertical
dy = voy*T + 0.5*ay*T^2
dy = vi*sin60*x/vi*cos60 - 0.5*g*x^2/(vi^2*(cos60)^2)
dy = tan60*x - 0.5*9.8*x^2/(vi^2*(cos60)^2)
(5)
x = 2 + 4 = 6m
dy = tan60*6 - 0.5*9.8*6^2/(10^2*(cos60)^2)
dy = 3.34 m
h = 3.34-3 = 0.34 m above the level of B
OPTION d
(6)
as the object lands on B
vertical distance dy = 3 m
3 = tan60*x - 0.5*9.8*x^2/(10^2*(cos60)^2)
x = 6.47 m from the starting point
6.47m - 6 = 0.47 m beyond the near side of building B
OPTION (e)
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