Answer :
Answer:
a) 17.40 s
b) 417.6 m
c) 6.04 times farther
Explanation:
a) In the maximum height vy = 0, so:
vy = v0y + at
For the upthorwn, a= -g, because the movement occurs against free-fall, so the up time can be calculated:
0 = v0.sinα - g*t
α = 31º and sin31º = 0.515
g = (1/6)*10 = 1.66 m/s²
1.66t = 28*0.515
t = 8.70 s
The uptime is equal to downtime, so the total time is:
2x8.70 = 17.40 s
b) The distance that it travels is the x-axis, so:
D = vx*t, where t is the total time 17.40s
D = v0*cosα*t
D = 28*cos31º*17.40
D = 417.6 m
c) The uptime on Earth would be:
10t = 28*0.515
t = 1.44 s
The total time would be: 2.88s
So, the distance on Earth:
De = 28*cos31º*2.88
De = 69.12 m
So
D/De = 417.6/69.12
D/De = 6.04 times farther
a) The ball was in flight for 17.40 s
b) The ball travels 417.6 m
c) It would have traveled 6.04 times farther on the moon than on earth.
Given information:
initial velocity u = 28m/s
the angle of projection θ = 31°
acceleration due to gravity on the moon [tex]g\approx\frac{10}{6}m/s^2[/tex]
(a) at the maximum height final velocity [tex]v_y = 0[/tex],
where subscript y denotes vertical direction.
[tex]v_y = u_y - gt[/tex]
[tex]0 = usin\theta - gt\\\\usin31=gt\\\\1.66t = 28*0.515\\\\t = 8.70 s[/tex]
total time of flight is T = 2t = 17.40s
(b) The distance traveled
[tex]d =u_x \times T, \\d = u\times cos31\times T\\d = 28\times cos31\times 17.40\\ d= 417.6 m[/tex]
(c) The time taken to reach the maximum height on Earth would be:
[tex]10t = 28\times 0.515\\t = 1.44 s[/tex]
The total time of flight would be T' = 2.88s
So, the distance on Earth:
[tex]d' = 28\times cos31\times2.88\\d' = 69.12 m[/tex]
the ratio of the distances:
d/d' = 417.6/69.12
d/d' = 6.04
Learn more about acceleration due to gravity:
https://brainly.com/question/17331289?referrer=searchResults