Asked by Anonymous
Two objects are thrown simultaneously from the same height at 45° angles to the vertical with a
speed of 20 m per second; one up, the other one down. Find the difference between the heights the
objects will be at two seconds later. How are these objects moving in regards to one another?
speed of 20 m per second; one up, the other one down. Find the difference between the heights the
objects will be at two seconds later. How are these objects moving in regards to one another?
Answers
Answered by
drwls
Y1 = 20 sin45*t - (g/2)t^2
Y2 = -20 sin45*t -(g/2)t^2
Y1 - Y2 = 40 sin45*t = 56.6 m apart
V1 = 20 sin45 - gt
V2 = -20 sin45 - gt
The vertical velocity difference between them remains
V2 - V1 = -40 sin 45 = -28.3 m/s
Two seconds after release, both are going down, but V2 is going down faster.
Y2 = -20 sin45*t -(g/2)t^2
Y1 - Y2 = 40 sin45*t = 56.6 m apart
V1 = 20 sin45 - gt
V2 = -20 sin45 - gt
The vertical velocity difference between them remains
V2 - V1 = -40 sin 45 = -28.3 m/s
Two seconds after release, both are going down, but V2 is going down faster.
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