Solving Maximum Range Throw for Astronaut on Earth and Planet with g1

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The discussion focuses on calculating the maximum range of a ball thrown by an astronaut on Earth and on a planet with a gravitational acceleration of g1 = 19 m/s². The optimal angle for maximum range is established as 45 degrees, and the necessary speed to achieve this range on Earth is derived from the equation v = sqrt(78.48), resulting in approximately 8.85 m/s. For the planet with g1 = 19 m/s², the maximum throw distance is calculated using the formula xp = 9.81 * 8 / 19, yielding a distance of approximately 4.13 m. The maximum height reached during this trajectory requires further calculation, utilizing the formula v² = v1² + 2ad, with the initial vertical velocity component being v1 = v * sin(45).

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Q. An astronaut in his space suit can throw a ball a maximum distance dmax = 8 m on the surface of the earth.


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a) For a given speed of the ball, what angle to the horizontal q (in degrees) will yield the greatest range?
q = ° 45 *


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b) If the ball is thrown at this same angle q, what speed will produce this greatest range (8 m) ?
v = m/s *
sqrt (78.48) OK


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c) How far can he throw the ball on a planet where g1 = 19 m/s2?
xp = m *
9.81*8/19 OK


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d) What height will the ball reach on this "maximum range" trajectory? (on the planet where g1 = 19 m/s2)
hmax = m
3.91 NO

Need help with part d,

I think we need to use the formula , X = V sqrt (2H/g), if yes, what to plug in,

Pl. Help
 
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for d
Vertical ONLY
what is your initial velocity COMPONENT that points upward? that's v1. What is the velocity of the ball at it's maximum height? what is the acceleration?
use this formula
[tex]v_{2}^2 = v_{1}^2 + 2ad[/tex]
and don't forget about a sign convention, take one direction (up or down) to be positive and the other to be negative. You cannot take the square root of a negative number - you have toi get a REAL number
 
At the maximum height, v2 = 0

so, 2ad = - v1^2

d = -v1^2/2(-9.81)

v1= v1sin(45)

I get 1.027m
 

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