Question
An astronaut on the planet Vimrom IV finds that she can jump a maximum horizontal distance of 12.00 m if her initial speed is 4.90 m/s. What is the acceleration due to gravity on this planet?
In the previous problem, what is the maximum height about the ground reached by the astronaut?
In the previous problem, what is the maximum height about the ground reached by the astronaut?
Answers
Damon
Do you know 45 degrees gives max distance?
u = 4.9 cos 45 = .707 (4.9) = 3.46 m/s
12 = 3.46 t
t = 3.47 seconds in air
t going up = 3.47/2 = 1.73 s upward
v = Vi - g t
Vi = 4.9 sin 45 = 3.46
v = 0 at top
0 = 3.46 - g (1.73)
g = 2 m/s^2
max height is average speed up time time up
= (3.46/2) * 1.73 = 3 meters
u = 4.9 cos 45 = .707 (4.9) = 3.46 m/s
12 = 3.46 t
t = 3.47 seconds in air
t going up = 3.47/2 = 1.73 s upward
v = Vi - g t
Vi = 4.9 sin 45 = 3.46
v = 0 at top
0 = 3.46 - g (1.73)
g = 2 m/s^2
max height is average speed up time time up
= (3.46/2) * 1.73 = 3 meters