Astronaut mark watney is very sick of potatoes. therefore, he throws a potato ata 30◦ angle above the horizontal as hard as he can. the initial height of the potatowhen it leaves his hand is 2.0 meters, and its initial velocity is 5.0 m/s (about 10 mph).gravitational acceleration on the surface of mars is 3.8 m/s2. how far will watneyhave to walk when he goes to retrieve the potato so he doesnt starve? (assume thatthe ground watney is standing on is flat and level.)

S = 8.132 m

Explanation:

∅ = 30°

Vi = 5 m/s

a = g = 3.8 m/s

we need to find horizontal and vertical component of velocity.

Vx= Vi cos∅

Vx= 5 * cos 30  

Vx= 4.33 m/s

Vy= Vi sin∅

Vy= 5* sin 30

Vy= 2.5 m/s

As horizontal component of velocity will remain constant. The distance covered by potato will be:

S = Vx * t

we need to find t ( time of flight):

Time required for potato to reach maximum height:

( we will use only vertical component of velocity to find time of flight)

Vfy= Viy - g*t                                   ( Viy= initial vertical velocity )

g * t = Viy                                         ( Vfy= 0 at maximum height)

t₁ = Viy / g = 2.5 / 3.8   sec  

t₁= 0.658 s

Height reached:

- 2gh₁ = Vfy² - Viy²                           ( Viy= initial vertical velocity )

                                                         ( Vfy= final vertical velocity )

- 2 * 3.8 * h = ( 0 )² - (2.5 m/s)²      ( ∴ Viy = 2.5 m/s & Vfy = 0 m/s)

 7.6 * h₁ = 6.25

⇒ h₁ = 6.25 / 7.6

⇒ h₁ = 0.822 m

Time required to reach the ground = t₂ = ?

First find total height = H = 2 m + 0.822 m     ( The potato is thrown from 2 m height )

H= 2.822 m

Now

H = Viy * t₂ + gt₂²

2.822= 0 * t₂ + *( 3.8* t₂²

2.822= 1.9 * t₂²

⇒ t₂² = 2.822/1.9 sec² = 1.485 sec²

⇒ t₂ =

⇒ t₂ = 1.22 s

Total time of flight = t = t₁ + t₂

t = 0.658 s + 1.22 s

t = 1.878 s

Now

S = Vx * t

S= 4.33 * 1.878

S = 8.132 m

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