Answer:
15.3 s and 332 m
Explanation:
With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon
gm = 1/6 ge
gm = 1/6 9.8 m/s² = 1.63 m/s²
We calculate the range
R = Vo² sin 2θ / g
R = 25² sin (2 30) / 1.63
R= 332 m
We will calculate the time of flight,
Y = Voy t – ½ g t2
Voy = Vo sin θ
When the ball reaches the end point has the same initial height Y=0
0 = Vo sin t – ½ g t2
0 = 25 sin (30) t – ½ 1.63 t2
0= 12.5 t – 0.815 t2
We solve the equation
0= t ( 12.5 -0.815 t)
t=0 s
t= 15.3 s
The value of zero corresponds to the departure point and the flight time is 15.3 s
Let's calculate the reach on earth
R2 = 25² sin (2 30) / 9.8
R2 = 55.2 m
R/R2 = 332/55.2
R/R2 = 6
Therefore the ball travels a distance six times greater on the moon than on Earth
You could be lying completley still on your bed, and all though it seems you are at rest, you are moving along with the earth around the sun and hence are motion. This is why 'being at rest' is more of a relative term. Hope this helps!
Answer:
a = 3 m/s^2
Explanation:
Vi = 10 m/s
Vf = 40 m/s
t = 10 s
Plug those values into the following equation:
Vf = Vi + at
40 = 10 + 10a
---> a = 3 m/s^2
I need someone to help me with my finals!!!!
