The length of the mercury thread is found to be 4cm and 24cm at ice point and steam point respectively on an ungraduated thermom
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Answer:
Explanation:
Let v is the launch speed of the plastic ball and the angle of projection is θ.
So, in horizontal direction
v Cosθ x t = 4.8 .... (1)
In th evertical direction
1.4 = v Sin θ x t - 0.5 gt² .... (2)
As , v Sin θ x t = 3.8 .... (3) , put in equation (2)
1.4 = 3.8 - 4.9 t²
t = 0.7 s
Put in (1) and (3)
v Cosθ x 0.7 = 4.8
v Cosθ = 6.86
and v Sinθ x 0.7 = 3.8
v Sinθ = 5.43
Now
v = 8.75 m/s