Answer:
15.34 s
332.1 m
6 times
Explanation:
g = (9.81 m/s^2) / 6 = 1.63 m/s^2
Since the ball is affected only by gravity, we can assume this to be a constant acceleration and we can use this equation on the vertical component:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
Y0 will be zero in this case because the ball was launched from the surface
a = -1.63 m/s^2 because gravity points down
Vy0 can be calculated as
Vy0 = V0 * sin(a) = 25 m/s * sin(30) = 12.5 m/s
Y(t) will equal zero in two points, when the ball was initially hit and when it hits the surface again:
0 = Vy0 * t + 1/2 * a * t^2
0 = t * (Vy0 + 1/2 * a * t)
t1 = 0 as expected, this is when the ball was hit
0 = Vy0 + 1/2 * a * t2
1/2 * a * t2 = -Vy0
t2 = -2 * Vy0/a
t2 = -2 * 12.5/(-1.63) = 15.34 s
The difference t2 - t1 = 15.34 s is how long the ball was in flight.
The distance it travelled is calculated with the horizontal speed.
Vx0 = V0 * cos(a) = 25 * cos(30) = 21.65 m/s
Since it is not affected by any horizontal forces, the horizontal speed remains constant. We can use the constant speed equation:
X(t) = X0 + Vx0 * t
X0 can be consifered zero if we deicide to put our origin of coordinates there.
X(15.34) = 21.65 * 15.34 = 332.1 m
X(15.34) - X0 = 332.1 m is how far the ball travelled.
On Earth it would have been affected by a larger gravity.
The time would have been
t2 = -2 * Vy0/a
t2 = -2 * 12.5/(-9.81) = 2.55 s
And travelled
X(15.34) = 21.65 * 2.55 = 55.2 m
It would travel 332.1/55.2 = 6 times further on the Moon than on Earth.
Answer:
15 s
32. 1 m
1/6 times
Explanation:
Thinking process:
The acceleration due to gravity = [tex]\frac{1}{9}* 9.81 m/s^{2}[/tex]
= 1.635 m/s²
Solving for the distance gives
s = ut - 1/2at²
= 33.1 m
Time taken will be:
[tex]t_{2}= 2v\\t_{2} = 2(12.5)(9.81)\\ = 2.55 s[/tex]
The distance traveled = vt
= [tex]21.65 (2.55)\\= 55.2[/tex]
Therefore, distance traveled on the surface =
