( If a person can jump a maximum horizontal distance (by using a 45° projection angle) of 1.33 m on Earth, what would be his maximum range on the Moon, where the free-fall acceleration is g/6 and g = 9.80 m/s2?
Let us assume the shape of the parabolic path will be the same.
Then compare heights reached
v at top = 0 = Vi - g t
so t = Vi/g
so t on moon = 6 * t on earth
Right here you can say it will spend six times as long above soil so goes six times as far since horizontal velocity is the same for both and constant.
h = Vi t - (1/2) g t^2
hmoon = 6 Vi t -(1/2)(g/6)(36 t^2)
or
hmoon = 6 Vi t - 3 g t^2
but t = Vi/g
so h = Vi^2/g -.5 g Vi^2/g^2
h = .5 Vi^2/g
h moon = 3 Vi^2/g
so
height moon = 6 * height earth and distance moon = 6 times distance earth
A 2.00 m tall basketball player is standing on the floor 10.0 m from the basket, as in the figure. If he shoots the ball at a 40.0° angle with the horizontal, at what initial speed must he throw the basketball so that it goes through the hoop without striking the back board? The height of the basket is 3.05 m
To find out the maximum range on the Moon, we can use the principles of projectile motion. The range of a projectile (horizontal distance traveled) can be calculated using the formula:
Range = (2 * v^2 * sinθ * cosθ) / g
Where:
- v is the initial velocity of the projectile
- θ is the angle of projection
- g is the acceleration due to gravity
Given that the initial velocity on both Earth and the Moon is the same (as the same person is jumping), and the angle of projection is 45°, we need to find the ratio between the accelerations due to gravity on Earth and the Moon.
On Earth, the acceleration due to gravity (g) is 9.80 m/s^2. On the Moon, the free-fall acceleration (g_moon) is given as g/6.
To find the ratio, we divide the acceleration due to gravity on the Moon by the acceleration due to gravity on Earth:
g_moon / g = (g/6) / g = 1/6
So, the acceleration due to gravity on the Moon is 1/6th of that on Earth.
Now, let's substitute the values into the range formula:
Range_moon = (2 * v^2 * sinθ * cosθ) / g_moon
Since we know that the maximum horizontal distance (range) on Earth is 1.33 m, we can substitute this value into the formula:
1.33 = (2 * v^2 * sin(45°) * cos(45°)) / g
Simplifying further, we can substitute the value for the ratio of accelerations:
1.33 = (2 * v^2 * (1/√2) * (1/√2)) / (9.8/6)
Now, solving for v^2:
v^2 = (1.33 * (9.8/6) * √2 * √2) / 2
v^2 = (1.33 * 9.8 * 6) / 6
v^2 = 1.33 * 9.8 ≈ 13.034
Taking the square root of both sides:
v ≈ √13.034 ≈ 3.611 m/s
Now we have the value for the initial velocity (v) on the Moon. We can substitute this into the range formula for the Moon:
Range_moon = (2 * (3.611)^2 * sin(45°) * cos(45°)) / (9.8/6)
Calculating further:
Range_moon = (2 * 13.034 * (1/√2) * (1/√2)) / (9.8/6)
Range_moon = 2 * (13.034 * 1/2) / (9.8/6)
Range_moon = (6.517 / 1.633) * 6
Range_moon ≈ 6.517 * 6 / 1.633
Range_moon ≈ 24.03 / 1.633 ≈ 14.717 m
Therefore, the maximum range on the Moon would be approximately 14.717 meters.