Projectile motion astronaut problem

Click For Summary
SUMMARY

The discussion centers on calculating the maximum height of a golf ball struck by an astronaut on the moon, where gravitational acceleration (g) is 1.6 m/s². The ball is launched at a velocity of 32 m/s at an angle of 35 degrees above the horizontal. The calculations reveal that the vertical component of the initial velocity (Vy) is 18 m/s, leading to a maximum height of 189 meters before descending into a crater 15 meters below the launch level.

PREREQUISITES
  • Understanding of projectile motion principles
  • Knowledge of trigonometric functions, specifically sine
  • Familiarity with kinematic equations
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation of projectile motion equations
  • Learn about the effects of different gravitational forces on projectile trajectories
  • Explore advanced kinematic problems involving multiple dimensions
  • Investigate the impact of air resistance on projectile motion
USEFUL FOR

Students studying physics, educators teaching projectile motion concepts, and anyone interested in the practical applications of kinematics in low-gravity environments.

Morhas
Messages
22
Reaction score
0

Homework Statement


An astronaut on the moon, where g=1.6m/s/s, strikes a golfball giving the ball a velocity of 32m/s [35degrees above the moons horizontal]. The ball lands in a crater floor that is 15m below the level where it was struck. Determine:

a) The maximum height of the ball


Homework Equations





The Attempt at a Solution



sin(35)=Vy/32
Vy= 18m/s

d=V1t+1/2at^2
0=18t + 0.5(-1.6)t^2
0.8t=18
t=22.5

d=18(22.5/2)+0.5(-1.6)(22.5/2)
d=198-8.8
d=189m
 
Physics news on Phys.org
Hi Morhas,

Morhas said:

Homework Statement


An astronaut on the moon, where g=1.6m/s/s, strikes a golfball giving the ball a velocity of 32m/s [35degrees above the moons horizontal]. The ball lands in a crater floor that is 15m below the level where it was struck. Determine:

a) The maximum height of the ball


Homework Equations





The Attempt at a Solution



sin(35)=Vy/32
Vy= 18m/s

d=V1t+1/2at^2
0=18t + 0.5(-1.6)t^2
0.8t=18
t=22.5

d=18(22.5/2)+0.5(-1.6)(22.5/2)

If you compare this to the equation you have above, you can see the time is supposed to be squared in the second term on the right hand side.

d=198-8.8
d=189m
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 22 ·
Replies
22
Views
5K
  • · Replies 23 ·
Replies
23
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
6K
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
5K
Replies
25
Views
3K
  • · Replies 11 ·
Replies
11
Views
6K
Replies
8
Views
4K