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

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An astronaut can throw a ball a maximum distance of 8 meters on Earth, achieving this at a 45-degree angle for optimal range. To reach this distance, the required speed of the ball is calculated using the formula for projectile motion, resulting in a speed of approximately 8.86 m/s. On a planet with a gravitational acceleration of 19 m/s², the astronaut can throw the ball a distance of about 4.14 meters. For the maximum height reached during this throw, the calculations involve determining the vertical component of the initial velocity and applying the appropriate kinematic equations. The final result for the maximum height on the planet is approximately 1.027 meters.
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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
v_{2}^2 = v_{1}^2 + 2ad
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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