Asked by Holly
Bryan tosses a 75.0 g rock into the lake. The rock leaves hus hand at a height of 175 cm above the ground travelling at 19.0 m/s.
How high above the ground will the rock travel if it is accidentally thrown straight up?
How high above the water will it travel if it is thrown in an arc and has a speed of 8.00 m/s at its maximum height?
How high above the ground will the rock travel if it is accidentally thrown straight up?
How high above the water will it travel if it is thrown in an arc and has a speed of 8.00 m/s at its maximum height?
Answers
Answered by
Henry
h = ho + (V^2-Vo^2)/2g
h = 1.75 + (0-19^2)/19.6 = 20.17 m.
Xo = 8 m/s = Hor. component of initial
velocity.
Yo = Ver. component of initial velocity.
Vo = 19 m/s = Total initial velocity.
cos A = Xo/Vo = 8/19 = 0.42105
A = 65.1o = Angle at which rock is thrown.
Yo = Vo*sin A = 19*sin 65.1 = 17.23 m/s.
h = 1.75 + (0-17.23^2)/-19.6 = 16.9 m.
h = 1.75 + (0-19^2)/19.6 = 20.17 m.
Xo = 8 m/s = Hor. component of initial
velocity.
Yo = Ver. component of initial velocity.
Vo = 19 m/s = Total initial velocity.
cos A = Xo/Vo = 8/19 = 0.42105
A = 65.1o = Angle at which rock is thrown.
Yo = Vo*sin A = 19*sin 65.1 = 17.23 m/s.
h = 1.75 + (0-17.23^2)/-19.6 = 16.9 m.
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