A) 2.64t
B) 2.64h
C) 2.64D
Explanation:
A)
The motion of the athlete is equivalent to the motion of a projectile, which consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- A uniformly accelerated motion (constant acceleration) along the vertical direction
The time of flight of a projectile can be found from the equations of motion, and it is found to be
[tex]t=\frac{2u sin \theta}{g}[/tex]
where
u is the initial speed
[tex]\theta[/tex] is the angle of projection
g is the acceleration due to gravity
In this problem, when the athlete is on the Earth, the time of flight is [tex]t[/tex].
When she is on Mars, the acceleration due to gravity is:
[tex]g'=0.379 g[/tex]
where g is the acceleration due to gravity on Earth. Therefore, the time of flight on Mars will be:
[tex]t'=\frac{2usin \theta}{g'}=\frac{2u sin \theta}{0.379g}=\frac{1}{0.379}t=2.64t[/tex]
B)
The maximum height reached by a projectile can be also found using the equations of motion, and it is given by
[tex]h=\frac{u^2 sin^2\theta}{2g}[/tex]
where
u is the initial speed
[tex]\theta[/tex] is the angle of projection
g is the acceleration due to gravity
In this problem, when the athlete is on the Earth, the maximum height is [tex]h[/tex].
When she is on Mars, the acceleration due to gravity is:
[tex]g'=0.379 g[/tex]
where g is the acceleration due to gravity on Earth. So, the maximum height reached on Mars will be:
[tex]h'=\frac{u^2 sin^2\theta}{2g'}=\frac{u^2 sin^2\theta}{(0.379)2g}=\frac{1}{0.379}h=2.64h[/tex]
C)
The horizontal distance covered by a projectile is also found from the equations of motion, and it is given by
[tex]D=\frac{u^2 sin(2\theta)}{g}[/tex]
where:
u is the initial speed
[tex]\theta[/tex] is the angle of projection
g is the acceleration due to gravity
In this problem, when the athlete is on the Earth, the horizontal distance covered is D.
When she is on Mars, the acceleration due to gravity is:
[tex]g'=0.379 g[/tex]
where g is the acceleration due to gravity on Earth. Therefore, the horizontal distance reached on Mars will be:
[tex]D'=\frac{u^2 sin(2\theta)}{g'}=\frac{u^2 sin(2\theta)}{(0.379)g}=\frac{1}{0.379}D=2.64D[/tex]
