Hon due to gravity g 9.80 m/S Useful Equations: x = xo + voxt +azt² vž = vổx + 2ax(x – x) Vx = Vox + axt X = - Xo +;(vox + Vx)t 1. (10 pts) A stone is thrown with the velocity vo = 10m/s at an angle of a = 40° to the horizon. (Disregard air resistance). Find: 1) the height which the ball will rise to. 2) the distance from the point of throwing to where the ball will drop onto the ground. 3) the time during which the ball will be in motion.
Hon due to gravity g 9.80 m/S Useful Equations: x = xo + voxt +azt² vž = vổx + 2ax(x – x) Vx = Vox + axt X = - Xo +;(vox + Vx)t 1. (10 pts) A stone is thrown with the velocity vo = 10m/s at an angle of a = 40° to the horizon. (Disregard air resistance). Find: 1) the height which the ball will rise to. 2) the distance from the point of throwing to where the ball will drop onto the ground. 3) the time during which the ball will be in motion.
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Transcribed Image Text:Useful Number: Acceleration due to gravity g= 9.80 m/s²
Useful Equations:
Vx = Vox + axt
x = Xo + Voxt +azt?
꽃=D v6x + 2ax(x-Ko)
X =
1
Xo +;(Vox + Vx)t
40° to the horizon. (Disregard air
1. (10 pts) A stone is thrown with the velocity vo = 10m/s
resistance). Find:
1) the height which the ball will rise to.
2) the distance from the point of throwing to where the ball will drop onto the ground.
3) the time during which the ball will be in motion.
an angle of a
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