Answer to Question #151020 in Physics for Arvin Shopon

Question #151020
The stone is thrown straight up from a height of 1.2 m at an initial speed of 11 m/ s. At what height and at what speed does the stone move 1.5 s after the moment of throwing?
1
Expert's answer
2020-12-14T12:16:25-0500

a) We can find the height of the stone 1.5 s after the moment of throwing from the kinematic equation:


y=y0+v0t12gt2,y = y_0 + v_0t-\dfrac{1}{2}gt^2,y=1.2 m+11 ms1.5 s129.8 ms2(1.5 s)2=6.67 m.y = 1.2\ m + 11\ \dfrac{m}{s}\cdot 1.5\ s-\dfrac{1}{2}\cdot 9.8\ \dfrac{m}{s^2}\cdot (1.5\ s)^2=6.67\ m.

b) We can find the speed of the stone 1.5 s after the moment of throwing from the kinematic equation:


v=v0gt=11 ms9.8 ms21.5 s=3.7 mv=v_0-gt=11\ \dfrac{m}{s}-9.8\ \dfrac{m}{s^2}\cdot 1.5\ s=-3.7\ m

The sign minus means that the speed of the stone directed downward.

Answer:

a) y=6.67 m.y=6.67\ m.

b) v=3.7 m,v=3.7\ m, downward.


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