InterviewSolution
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InterviewSolution
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An object is projected upward with a 30^(@) launch angle and an initial speed of 60 m//s. How many second will it be in the air ? How far will it travel horizontally ? |
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Answer» Solution :The total time the object spend in the air is equal to twice the time REQUIRED to reach the top of the trajectory (because the parabola is symmetrical ). We can FIND the time required to reach the top by setting `v_(y)` equal to 0 and then doubling that amount of time : ` v_(y) overset("set")= 0 IMPLIES v_(0_(y))+(-g) t= 0` ` t= (v_(0_(y)))/(g)=(v_(0) sin theta_(0))/(g)=((60 m//s) sin 30^(@))/(10 m//s)=3 s` Therefore, the total flight time (that is, up and down) is `T= 2t = 2xx (3s)=6s`. Now , using the first horizontal- MOTION equation, we can calculate the horizontal displacement after 6 seconds. `Delta x = v_(0_(x))T=(v_(0) cos theta) T= [(60 m//s) cos 30^(@)](6s)=310 m` By the way, the full horizontal displacement of a projectile is CALLED the projectile's range. |
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